UC-NRLF EH MEMOEIAM Edward Bright (Brcenlcafs Jttathematical 3ttk^ UNIVERSITY ALGEBRA. DESIGNED FOR THE USE OF SCHOOLS AND COLLEGES. PREPARED BY WEBSTER WELLS, S. B., i ASSISTANT PROFESSOR OF MATHEMATICS IN THE MASSACHUSETTS INSTITUTE OF TECHNOLOGY; LEACH, SHEWELL, AND SANBORN, BOSTON AND NEW YORK. COPYRIGHT, 1880. q/v- By WEBSTER WELL?.. PREFACE. This work was designed to take the place of Green- leaf's Higher Algebra, portions of which have been used in the preparation of the present volume. It contains the topics usually taught in High Schools and Colleges, and the author's aim has been to present the subject in a compact form and in clear and concise language. The principles have been developed with regard to logical ac- curacy, and care has been given to the selection of exam- ples and practical illustrations which should exercise the student in all the common applications of the algebraic analysis. The full treatment given in the earlier chap- ters renders the previous study of a more elementary text-book unnecessary. Attention is invited to the following chapters, including those in which the most important changes have been made in the Higher Algebra : — Parentheses. Factoring. Zero and Infinity. Theory of Exponents. Simultaneous Equations involving Quadratics. Binomial Theorem for Positive Integral Exponents. Undetermined Coefficients. Logarithms. The answers have been put by themselves in the back part of the book, and those have been omitted which, if 7979-5 i / PREFACE. given, would destroy the utility of the problem. The ex- amples are over eighteen hundred in number, and are pro- gressive, commencing with simple applications of the rules, and passing gradually to those which require some thought for their solution. The works of Todhunter and Hamblin Smith, and other standard volumes, have been consulted in the preparation of the work, and have furnished a number of examples and problems. The author has also received numerous suggestions from practical teachers, to whom he would here express his thanks. WEBSTER WELLS. Boston, 1884. UNIVERSITY ALGEBRA. CONTENTS. Chapter Paob I. Definitions and Notation 1 Symbols of Quantity 1 Symbols of Operation 2 Symbols of Relation 4 Symbols of Abbreviation 5 Algebraic Expressions 5 Axioms 8 Negative Quantities 12 II. Addition 14 III. Subtraction 19 IV. Use of Parentheses 21 V. Multiplication 24 VI. Division 31 VII. Formula 38 VIII. Factoring 40 IX. Greatest Common Divisor 53 X. Least Common Multiple 61 XI. Fractions 66 Reduction of Fractions 70 Addition and Subtraction of Fractions 78 Multiplication of Fractions 82 Division of Fractions 85 Complex Fractions 87 XII. Simple Equations. — One unknown quantity 89 Transformation of Equations 91 Solution of Equations 95 Vlll CONTENTS. XIII. Problems. — One unknown quantity 103 XIV. Simple Equations. — Two unknown quantities 113 Elimination 115 XV. Simple Equations. — More than two unknown quan- tities 123 XVI. Problems. —More than one unknown quantity 127 Generalization of Problems 133 XVII. Discussion of Problems 137 Interpretation of Negative Results 139 XVIII. Zero and Infinity 142 Problem of the Couriers 143 XIX. Inequalities 148 XX. Involution 153 Involution of Monomials 153 Involution of Polynomials 154 Square of a Polynomial 155 Cube of a Binomial 156 Cube of a Polynomial 157 XXI. Evolution 158 Evolution of Monomials 159 Square Root of Polynomials 160 Square Root of Numbers 163 Cube Root of Polynomials 168 Cube Root of Numbers 171 Any Root of Polynomials 174 XXII. The Theory of Exponents 176 XXIII. Radicals 188 Reduction of Radicals 1 88 Addition and Subtraction of Radicals 193 Multiplication of Radicals 194 Division of Radicals 196 Involution of Radicals 197 Evolution of Radicals 198 Reduction of Fractions with Irrational Denominators 199 Imaginary Quantities 202 Quadratic Surds 205 Radical Equations 208 CONTENTS. ix XXIV. Quadratic Equations. — One unknown quantity... 210 Pure Quadratic Equations 211 Affected Quadratic Equations 213 XXV. Problems. — Quadratic Equations. — One unknown quantity 223 XXVI. Equations in the Quadratic Form 227 XXVII. Simultaneous Equations involving Quadratics 233 XXVIII. Problems. — Quadratic Equations. — Two unknown QUANTITIES 244 XXIX. Theory of Quadratic Equations 249 Discussion of the General Equation 249 XXX. Discussion of Problems leading to Quadratic Equations 257 Interpretation of I maginary Results 259 Problem of the Lights 259 XXXI. Ratio and Proportion 262 XXXII. Variation 270 XXXIII. Arithmetical Progression 274 XXXIV. Geometrical Progression 282 XXXV. Harmonical Progression 291 XXXVI. Permutations and Combinations 294 XXXVII. Binomial Theorem. — Positive Integral Exponent 298 XXXVIII. Undetermined Coefficients 304 Expansion of Fractions into Series 307 Expansion of Radicals into Series 310 Decomposition of Rational Fractions 312 Reversion of Series 318 XXXIX. Binomial Theorem. —Any Exponent 321 XL. Summation of Infinite Series 328 Recurring Series 328 The Differential Method 332 Interpolation 336 x CONTENTS. XLI. Logarithms 339 Properties of Logarithms 342 Use of the Table 348 Solutions of Arithmetical Problems by Logarithms 354 Exponential Equations 358 Application of Logarithms to Problems in Compound In- terest 359 Exponential and Logarithmic Series 362 Arithmetical Complement 366 XLII. General Theory of Equations 369 Divisibility of Equations 370 Number of Roots .- 371 Formation of Equations 373 Composition of Coefficients 374 Fractional Roots 376 Imaginary Roots 376 Transformation of Equations 377 Descartes' Rule of Signs 383 Derived Polynomials 385 Equal Roots 386 Limits of the Roots of an Equation 389 Sturm's Theorem 392 XLIII. Solution of Higher Numerical Equations 399 Commensurable Roots 400 Recurring or Reciprocal Equations 404 Cardan's Method for the Solution of Cubic Equations. .... 408 Biquadratic Equations 411 Incommensurable Roots 412 Horner's Method 412 Approximation by Double Position 417 Newton's Method of Approximation 419 Answers to Examples 421 Table of the Logarithms of Numbers from 1 to 10,000. .Appendix. ALGEBRA. I. — DEFINITIONS AND NOTATION. 1. Quantity is anything that can be measured ; as dis- tance, time, weight, and number. 2. The Measurement of quantity is accomplished by find- ing the number of times it contains another quantity of the same kind, assumed as a standard. This standard is called the unit of measure. 3. Mathematics is the science of quantities and their re- lations. 4. Algebra is that branch of mathematics in which the relations of quantities are investigated, and the reasoning abridged and generalized, by means of symbols. 5. The Symbols employed in Algebra are of four kinds: symbols of quantity, symbols of operation, symbols of relation, and symbols of abbreviation. SYMBOLS OF QUANTITY. 6. The Symbols of Quantity generally used are the figures of Arithmetic and the letters of the alphabet. The figures are used to represent known quantities and determined values, and the letters any quantities whatever, known or unknown. 7. Known Quantities, or those whose values are given, ■2 ALGEBRA. when not expressed by figures, are usually represented by the first letters of the alphabet, as a, b, c. 8. Unknown Quantities, or those whose values are not given, are usually represented by the last letters of the alphabet, as x, y, z. 9. Zero, or the absence of quantity, is represented by the symbol 0. 10. Quantities occupying similar relations in different op- erations are often represented by the same letter, distinguished by different accents, as a', a", a'", read "a prime," "a second," "a third," etc. ; or by different subscript figures, as a 1} a 2) a s , read " a one/' " a two," " a three," etc. SYMBOLS OF OPERATION. 11. The Symbols of Operation are certain signs or char- acters used to indicate algebraic operations. 12. The Sign of Addition, + , is called "plus." Thus, a -(- b, read " a plus b" indicates that the quantity b is to be added to the quantity a. 13. The Sign of Subtraction, — , is called "minus." Thus, a — b, read " a minus b" indicates that the quantity b is to be subtracted from the quantity a. The sign ~ indicates the difference of two quantities when it is not known which of them is the greater. Thus, a ~ b indicates the difference of the two quantities a and b. 14. The Sign of Multiplication, x , is read " times" "into," or "multiplied by." Thus, aXb indicates that the quantity a is multiplied by the quantity b. A simple point (. ) is sometimes used in place of the sign X- The sign of multiplication is, however, usually omitted, except between two arithmetical figures separated by no other sign; multiplication is therefore indicated by the absence of any sign. Thus, 2ab indicates the same as 2 X a X b, or 2 . a . b. DEFINITIONS AND NOTATION. 3 15. The quantities multiplied are called factors, and the result of the multiplication is called the product. 16. The Sign of Division, ~, is read "divided by." Thus, a -4- b indicates that the quantity a is divided by the quantity b. Division is otherwise often indicated by writing the divi- dend above, and the divisor below, a horizontal line. Thus, - indicates the same as a -i- b. Also, the sign of division b may be replaced in an operation by a straight or curved line. Thus, a I b, or b ) a, indicates the same as a -f- b. 17. The Exponential Sign is a figure or letter written at the right of and above a quantity, to indicate the number of times the quantity is taken as a factor. Thus, in x 3 , the 3 in- dicates that x is taken three times as a factor ; that is, x 3 is equivalent to xxx. The product obtained by taking a factor two or more times is called a power. A single letter is also often called the first power of that letter. Thus, a 2 is read "a to the second power," or "a square," and indicates a a; a 3 is read " a to the third power," or " a cube," and indi- cates acta; a 4 is read " a to the fourth power," or " a fourth," and indi- cates a a a a; a n is read " a to the nth power," or " a nth," and indicates a a a etc., to n factors. The figures or letters used to indicate powers are called exponents ; and when no exponent is written, the first power is understood. Thus, a is equivalent to a 1 . The root of a quantity is one of its equal factors. Thus, the root of a 2 , a 3 , or a* is a. 18. The Radical Sign, \f , when prefixed to a quantity, indicates that some root of the quantity is to be extracted. 4 ALGEBRA. Thus, SJ a indicates the second or square root of a; ^]a indicates the third or cuhe root of a; $ a indicates the fourth root of a; and so on. The index of the root is the figure or letter written over the radical sign. Thus, 2 is the index of the square root, 3 of the cube root ; and so on. When the radical sign has no index written over it, the index 2 is understood. Thus, sj a is the same as tf a. SYMBOLS OF RELATION. 19. The Symbols of Relation are signs used to indicate the relative magnitudes of quantities. 20. The Sign of Equality, =, read " equals" or "equal to," indicates that the quantities between which it is placed are equal. Thus, x = y indicates that the quantity x is equal to the quantity y. A statement that two quantities are equal is called an equation. Thus, x -\- 4=2 x — 1 is an equation, and is read " x plus 4 equals 2x minus 1." 21. The Sign of Ratio, : , read " to," indicates that the two quantities between which it is placed are taken as the terms of a ratio. Thus, a : b indicates the ratio of the quan- tity a to the quantity b, and is read " the ratio of a to b." A proportion, or an equality of ratios, is expressed by writ- ing the sign =, or the sign : :, between equal ratios. Thus, 30 : 6 = 25 : 5 indicates that the ratio of 30 to G is equal to the ratio of 25 to 5, and is read "30 is to 6 as 25 is to 5." 22. The Sign of Inequality, > or < , read " is greater than" or "is less than" respectively, when placed between two quantities, indicates that the quantity toward which the opening of the sign turns is the greater. Tims, x > y is read "x is greater than y" ; x—6< y is read "x minus 6 is less than y." DEFINITIONS AND NOTATION. 5 23. The Sign of Variation, cc, read "varies as," indicates that the two quantities between which it is placed increase or diminish together, in the same ratio. Thus, a oc _ is read " a varies as c divided by d." SYMBOLS OF ABBREVIATION. 24. The Signs of Deduction, .-. and v , stand the one for therefore or hence, the other for since or because. 25. The Signs of Aggregation, the vinculum , the bar | , the parenthesis ( ) , the brackets [ ] , and the braces j £, indicate that the quantities connected or enclosed by them are to be subjected to the same operations. Thus, a + b X oc, a x, (a + b) x, [a + fr]x, \a + bc x, b all indicate that the quantity a + b is to be multiplied by x. 26. The Sign of Continuation, , stands for and so on, or continued by the some law. Thus, a, a + b, a + 2 b, a + 3 b, is read "a, a plus b, a plus 2b, a plus 3 b, and so on." ALGEBRAIC EXPRESSIONS. 27. An Algebraic Expression is any combination of alge- braic symbols. 28. A Coefficient of a quantity is a figure or letter pre- fixed to it, to show how many times the quantity is to be taken. Thus, in 4«, 4 is the coefficient of a, and indicates that a is taken four times, or a + a + a + a. Where any number of quantities are multiplied together, the product of 6 ALGEBEA. any of them may be regarded as the coefficient of the product of the others; thus, in abed, ab is the coefficient of ed, b of ac d, a b d of c, and so on. When no coefficient of a quantity is written, 1 is understood to he the coefficient. Thus, a is the same as 1 a, and x y is the same as 1 x y. 29. The Terms of an algebraic expression are its parts connected by the signs + or — . Thus, a and b are the terms of the expression a + b ; 2 a, b 2 , and — 2 a c, of the expression 2 a + b 2 — 2 a c. 30. The Degree .of a term is the number of literal factors which it contains. Thus, 2 a is of the first degree, as it contains but one literal factor, a & is of the second degree, as it contains two literal factors. 3 a b 2 is of the third degree, as it contains three literal factors. The degree of any term is determined by adding the expon- ents of its several letters. Thus, a b 2 c s is of the sixth degree. 31. Positive Terms are those preceded by a plus sign ; as, + 2 a, or + a b 2 . When a term has no sign written, it is understood to be positive. Thus, a is the same as + a. Negative Terms are those preceded by a minus sign ; as, — 3 a, or —be. This sign can never be omitted. 32. In a positive term, the coefficient indicates how many times the quantity is taken additively (Art. 28) ; in a nega- tive term, the coefficient indicates how many times the quan- tity is taken svbtractively. Thus, + 2 x is the same as + x + x ; — 3 a is the same as — a — a — a. DEFINITIONS AND NOTATION. 7 33. If the same quantity be both added to and subtracted from another, the value of the latter will not be changed ; hence if any quantity b be added to any other quantity a, and b be subtracted from the result, the remainder will be a; that is, # a + b — b = a. Consequently, equal terms affected by unlike signs, in an expression, neutralize each other, or cancel. 34. Similar or Like Terms are those which differ only in their numerical coefficients. Thus, 2 x y 2 and — 1 xif are similar terms. Dissimilar or Unlike Terms are those which are not similar. Thus, b x 2 y and bxy 2 are dissimilar terms. 35. A Monomial is an algebraic expression consisting of only one term ; as, 5 a, 7 a b, or 3 b' 2 c. A monomial is sometimes called a simple quantity. 36. A Polynomial is an algebraic expression consisting of more than one term ; as, a + b, or 3 a 2 + b — 5 b 3 . A polynomial is sometimes called a compound quantity, or a multinomial. 37. A Binomial is a polynomial of two terms ; as, a — b, 2 a + b 2 , ov dad 2 — b. A binomial whose second term is negative, as a — b, is some- times called a residual. 38. A Trinomial is a polynomial of three terms ; as, a + b + c, or a b + c 2 — b 3 . 39. Homogeneous Terms are those of the same degree ; as, a 2 , 3 be, and — 4 x 2 . $ ALGEBRA. 40. A polynomial is homogeneous when all its terms are homogeneous ; as, a 3 + 2 a b c — 3 b 3 . 41. A polynomial is said to be arranged according to the decreasing powers of any letter, when the term having the highest exponent of that letter is placed first, that having the next lower immediately after, and so on. Thus, a 3 +3a 2 b + 3ab 2 + b 3 is arranged according t6 the decreasing powers of a. A polynomial is said to be arranged according to the increas- ing powers of any letter, when the term having the lowest exponent of that letter is placed first, that having the next higher immediately after, and so on. Thus, a 3 + 3 a 2 b + 3 a b 2 + b 3 is arranged according to the increasing powers of b. 42. The Reciprocal of a quantity is 1 divided by that quantity. Thus, the reciprocal of a is - , and of x + y is a x + y 43. The Interpretation of an algebraic expression consists in rendering it into an arithmetical quantity, by means of the numerical values assigned to its letters. The result is called the numerical value of the expression. Thus, the numerical value of 4d+ 3 be — d when a = 4, b = 3, c = 5, and d = 2, is 4x4 + 3x3x5-2 = 16 + 45 - 2 = 59. AXIOMS. 44. An Axiom is a self-evident truth. Algebraic operations are based upon definitions, and the following axioms : — DEFINITIONS AND NOTATION. 9 1. If the same quantity, or equal quantities, be added to equal quantities, the sums will he equal. • 2. If the same quantity, or equal quantities, he subtracted from equal quantities, the remainders will he equal. 3. If equal quantities be multiplied by the same quantity, or by equal quantities, the products will be equal. 4. If equal quantities be divided by the same quantity, or by equal quantities, the quotients will be equal. 5. If the same quantity be both added to and subtracted from another, the value of the latter will not be changed. 6. If a quantity be both multiplied and divided by another, the value of the former will not be changed. 7. Quantities which are equal to the same quantity are equal to each other. 8. Like powers and like roots of equal quantities are equal. 9. The whole of a quantity is equal to the sum of all its parts. EXERCISES ON THE PRECEDING DEFINITIONS AND PRINCIPLES. 45. Translate the following algebraic expressions into ordinary language : d m 1. 3 a 2 + b c — q. 5. cd : — = ab : \J x*. 3 n x 2. 4 m . 6. (a — b)x = [c + d~] y. <- -} 3a — d " 2c + b 3. ^ a + b = ^ a 2 — c. 7. {m + r — s}n = 4,mn>pa. 8. \J -^- < (e - d) (h + fj. 46. Put into the form of algebraic expressions the follow- ing : 1. Five times a, added to two times b. 2. Two times x, minus y to the second power. 10 ALGEBRA. 3. The difference of x and y. 4. The product of «, b, c square, and d cube. 5. x + y multiplied by a — b. 6. a square divided by the sum of b and c. 7. x divided by 3, increased by 2, equals three times y, diminished by 11. 8. The reciprocal of a + b, plus the square of a } minus the cube root of b, is equal to the square root of c. 9. The ratio of 5 a divided by b, to d divided by c square, equals the ratio of x square y cube to y square z fourth. 10. The product of in and a + b is less than the reciprocal of x cube. 11. The product of x + y and x — y is greater than the product of the square of a — d into the cube of a + b. 12. The quotient of a divided by 3 a — 2 is equal to the square root of the quotient of m + n divided by 2x — y 2 . 47. Find the numerical values of the following : — When a = 6, b = 5, c = 4, and d — 1, of 1. a 2 + 2 a b — c + d. 4. a 2 (a + b) — 2abc. 2. 2a?-2a 2 b + c 3 . 5. 5a 2 b-±ab 2 + 21c. 3. 2a 2 + 3bc-5. 6. 7 a 2 + (a- b) (a-c). When a = 4, b = 2, c = 3, and d = l, of a 2 J,2 1.15a-7(b+c)-d. 10. ^ + | + ^. 8. 25a 2 -7(b 2 + c 2 ) + d\ 11. 4 + 1 . „« £» c '25 a — 30 c— d 9. - H 1- - . 12. . bed b + c When a = \, b = \, c = \, and x = 2, of IS. (2 a + 3 b + 5 c) ($ a + 3 b - 5 e) (2 a- 3 b + 15 c). DEFINITIONS AND NOTATION. H 15. x* - (2 a + 3 b) x 3 + (3 a - 2 b) x 2 - ex + be. When a=b, and b = £, of 16 5 a + ^ - C 3 a - ( 2 a - ^)] 17 13 a + 3 b + {7 Q + b) + [3 a + 8 (4 a - b)~\) 2a + Sb When b = 3, c = 4, d = 6, and e = 2, of 18. V 27T- v' 27+ y/2Z 19. V 3^7+ ^"9^-^27. When « = 16, & = 10 ; cc = 5, and y=l, of 20. (b - x) (y/a + b) + ^ (a - b) (x + y). 48. What is the coefficient of 1. x in 3 n 2 x ? 3. x y in — 20 m? xyz z ? 2. a c s in a J 2 c 3 d 4 ? 4. ra 2 w 3 in 5 a 8 m 2 a; ?i 3 ? What is the degree of 5. 3ax? 6. 2m*nx*? 7. a 2 b s c 2 d 5 ? 8. 2mcr 2 y 3 .?? Arrange the following expressions according to the increas- ing powers of x : 9. 2cc 2 -3a; + x 3 + l-4a; 4 . 10. 3 x y* — 5 x s y + y i — x 4 — x 2 y 2 . Arrange the following expressions according ' o the decreas- ing powers of a : 11. 1- a 2 -2 a + a 3 + 2 a*. 12. aJ 3 -i 4 + a 4 -4« 2 i 2 -3a 3 5. 12 ALGEBRA. NEGATIVE QUANTITIES. 49. The signs + and — , besides indicating the operations of addition and subtraction, are also used, in Algebra, to indi- cate the nature or quality of the quantities to which they are prefixed. To illustrate, let us suppose a person, having a property of $ 500, to lose $ 150, then gain $ 250, and finally to incur a debt of $ 450 ; it is required to find the amount of his property. Since gains have an additive effect on property, and debts or losses a subtractive effect, we may indicate these different qualities algebraically by prefixing the signs + and — to them, respectively ; thus, we should represent the transactions as follows, $ 500 - $ 150 + $ 250 - $ 450 ; which reduces to $ 150, the amount required. But suppose, having a property of $ 500, he incurs a debt of $ 700 ; we should represent the transaction algebraically as follows, $500-1700; or, as incurring a debt of $ 700 is equivalent to incurring two debts, one of $ 500 and the other of $ 200, the transaction may be expressed thus, $ 500 - $ 500 - $ 200. Now since, by Art. 33, $ 500 and — $ 500 neutralize each other, we have remaining the isolated negative quantity — $200 as the algebraic representative of the required prop- erty. In Arithmetic, we should say that he owed or was in debt $200; in Algebra, we make also the equivalent state- ment that his property amounts to —$200. In this way we can conceive the possibility of the indepen- dent existence of negative quantities; and as, in Arithmetic, losses may be added, subtracted, multiplied, etc., precisely as though they were gains, so, in Algebra, negative quantities DEFINITIONS AND NOTATION. 13 may be added, subtracted, multiplied, etc., precisely as though they were positive. The distinction of positive and negative quantities is applied in a great many cases in the language of every-day life and in the mathematical sciences. Thus, in the thermometer, we speak of a temperature above zero as +, and one below as — ; for instance, +25° means 25° above zero, and —10° means 10° below zero. In navigation, north latitude is considered -j-, and south latitude — ; longitude west of Greenwich is con- sidered +, and longitude east of Greenwich — ; for example, a place in latitude — 30°, longitude + 95°, would be in latitude 30° south of the equator, and in longitude 95° west of Green- wich. And, in general, when we have to consider quantities the exact reverse of each other in quality or condition, we may regard quantities of either quality or condition as posi- tive, and those of the opposite quality or condition as negative. It is immaterial which quality we regard as positive ; but hav- ing assumed at the commencement of an investigation a certain quality as positive, we must retain the same notation through- out. The absolute value of a quantity is the number represented by that quantity, taken independently of the sign affecting it. Thus, 2 and — 2 have the same absolute value. But as we consider a person who owns $ 2 as better off than one who owes $2, so, in Algebra, we consider + 2 as greater than — 2 ; and, in general, any positive quantity, however small, is considered greater titan any negative quan- tity. Also, as we consider a person who owes $ 2 as better off than one who owes $3, so, in Algebra, we consider — 2 as greater than —3; and, in general, oftivo negative quantities, that is regarded as the greater which has the less number of units, or which has the smaller absolute value. Again, as we consider a person who has no property or debt as better off than one who is in debt, so, in Algebra, zero is considered greater than any negative quantity. 14 ALGEBRA. II. — ADDITION. 50. Addition, in Algebra, is the process of collecting two or more quantities into one equivalent expression, called, the sum. 51. In Arithmetic, when a person incurs a debt of a certain amount, we regard his property as diminished by the amount of the debt. So, in Algebra, using the interpretation of nega- tive quantities as given in Art. 49, adding a negative quantity is equivalent to subtracting an equal positive quantity. Thus, the sum of a and — b is obtained by subtracting b from a, giv- ing as a result a — b. Hence, the addition of monomials is indicated by uniting the quantities with their respective signs. Thus, the sum of a, — b, c, d, — e, and — f, is a—b+c+d—e —f. The addition of polynomials is indicated by enclosing them in parentheses (Art. 25), and uniting the results with + signs. Thus, the sum of a + b and c — d is (a + b) + (c — d). 52. Let it be required to add c — d to a + b. If we add c to a + b, the sum will be a + b + c. But we have to add to a + b a quantity which is d less than c. Conse- quently our result is d too large. Hence the required sum will be a + b + c diminished by d, or a + b + c — d. Hence, the addition of polynomials may also be indicated by uniting their terms with their respective signs. 53. Let it be required to add 2 a and 3 a. By Art. 32, 2 a = a + a, and 3 a = a + a + a. ADDITION. 15 Hence (Art. 52) the sum of 2 a and 3 a is indicated by a -\- a + a + a + a, which, by Art. 32, is equal to 5 a. Hence, 2a + 3a = 5a. 54. Let it be required to add — 3 a and — 2 a. By Art. 32, — 3 a = — a — a — a, and — 2 a = — a — a. Hence (Art. 52), the sum of — 3 a and — 2 a is indicated by — a — a — a — a — a, or — 5 a (Art. 32). Hence, — 3 a — 2 a = — 5 a. From our ideas of negative quantities (Art. 49), we may ex- plain this result arithmetically as follows : If a person has two debts, one of $ 3 and the other of $ 2, he may be considered to be in debt to the amount of $ 5. 55. Let it be required to add 4 a and — 2 a. 4 a = a + a+ a + a, and — 2« = — a — a. Hence, the sum of 4 a and — 2 a is indicated by a + a + a + a — a — a. Now, by Art. 33, the third and fourth terms are neutralized by the fifth and sixth, leaving as the result a + a, or 2 a. Hence, 4 « — 2 a = 2 a. We may explain this result arithmetically as follows : If a person has $4 in money, and incurs a debt of $2, his property may be considered to amount to $ 2. 56. Let it be required to add — 4 a and 2 a. — 4:a = — a — a — a — a, and 2a = ffl+«. 16 ALGEBRA. Hence, the sum of — 4 a and 2 a is indicated by — a — a — a — a + a + a. The third and fourth terms neutralize the fifth and sixth, leaving as the result — a — a or —2a. Hence, — 4ta + 2a = — 2a. We may explain this result arithmetically as follows : If a person has $ 2 in money, and incurs a debt of 84, he may be considered to be in debt to the amount of $2. .57. From Arts. 55 and 56 we derive the following rule for the addition of two similar (Art. 34) terms of opposite sign: To add two similar terms, the one positive and the other negative, subtract the smaller coefficient from the larger, affix to the result the common symbols, and prefix the sign of the larger. For example, the sum of 7 x y and — 3xy is 4:xy, the sum of 3 a 2 b s and — 11 a~ b 3 is — 8 a 2 b 3 . 58. In Arithmetic, when adding several quantities, it makes no difference in which order we add them ; thus, 3 + 5 + 9, 5 + 3 + 9, 9+3 + 5, etc., all give the same result, 17. So also in Algebra, it is immaterial in what order the terms are united, provided each has its proper sign. Thus, — b + a is the same as a — b. Hence, in adding together any number of similar terms, some positive and some negative, we may add the positive terms first, and then the negative, and finally combine these two results by the rule of Art. 57. Thus, in finding the sum of 2 a, — a, la, 6 a, — 4 a, and — 5 a, the sum of the positive terms 2 a, 7 a, and 6 a, is 15 a, and the sum of the negative terms — a, — 4 a, and — 5 a, is — 10 a ; and the sum of 15 a and — 10 a is 5 a. 59. Let it be required to add 6 a — 7 x, 3 x — 2 a + 3 y, and 2 x — a — mn. ADDITION. 1 7 We might obtain the sum in accordance with Art. 52, by- uniting the terms by their respective signs, and combining similar terms by the methods previously given. It is however customary in practice, and more convenient, to set the expres- sions down one underneath the other, similar terms being in the same vertical column ; thus, 6 a — 1 x — 2 a + 3 x + 3 y — a + 2 x —mn 3 a — 2 x + 3 y — m n. It should be remembered that only similar terms can be combined by addition ; and that the algebraic sum of dissimilar terms can only be indicated by uniting them by their respective signs. 60. From the preceding principles and illustrations is de- rived the following RULE. To add together two or more expressions, set them down one underneath the other, similar terms being in the same vertical column. Find the sum of the similar terms, and to the result obtained unite the dissimilar terms, if any, by their respective signs. EXAMPLES. 1. 2. 3. 4. 5. la — 6 m 13 n — 4 a x 2a 2 b 3 a m n — 3 ax -a-b a — 11 m — 20n a x 11 a- b 5 a — 5m 6 n — lax -5a 2 b 11a — m 8n — ax ±a 2 b a 20 m — n 12 ax -9aH 18 ALGEBRA. 6. 7. 8. la — mp 2 2 a — 3 x ab + c d a + 6 mp 2 — a + 4x — ab -\- cd — 11 a — 3 mp 2 a + x 3 a b — 2 cd 8 a + 11 m j9 2 5 a — 7 a; lab — 5 cd — 9 a— 1 mp 2 —4: a— x —4tab + 6cd 18 a — 15 mp 2 —3 a + 1 x 2 ab — 5cd Find the sum of the following : 9. 4,xy z, — 3xy z, — 5xy z, 6x y z, — 9xy z, and 3 x y z. 10. 5 m n 2 — 8x 2 y, — m n 2 + x 2 y, — 6m n 2 — 3x 2 y, 4:mn 2 + 1 x 2 y, 2 m n 2 + 3 x 2 y, and — ra ri 2 — 2x 2 y. 11. 3a 2 + 2ab + 4,b 2 , 5a 2 -Sab + b 2 , -a 2 +5ab-b 2 , 18 a 2 -20 ab- 19 b 2 , and 14 a 2 - 3 a b + 20 b 2 . 12. 2a — 5b — c +1, 3b — 2-6a + 8c, c + 3a-4, and 1 + 2 b - 5 c. 13. 6x — 3y+lm, 2 n — x + y, 2 y — 4x— 5 m, and m + n — y. 14. 2 a - 3 b + 4 <7, 2 & - 3 d + 4 c, 2 cZ - 3 e + 4 a, and 2c-3a + 4i. 15. 3 cc — 2 y — z, 3 y — 5 x — 1 z, 8 z — y — x, and 4 x. 16. 2 m — 3n + 5r — t, 2 n — 6 t — 3 r — m, r + 3 m — 5n + 4t, and 3 t — 2 r + 1 n — 4 m. 17. 4:inn + 3 ab — 4 c, 3 x — 4 a b + 2 m n, and 3 m 2 — 4 p. 18. 3 a + b — 10, c — d — a, —4c + 2a — 3b — l, and 4 a; 2 + 5 - 18 m. 19. 4a;8_5 a 8_ 5ax 2 +6a 2 x, 6a s + 3x* + Aax 2 + 2a 2 x, -11 X s +19 ax 2 - 15 a 2 x, and 10 x 3 + 1 a 2 x + 5 a 3 - 18 a x 2 . 20. la — 5if, S^x + 2a, oif — \/x, and — 9a + 7tfx. 21. 3 a b + 3 (« + b), - a b + 2 (a + b), 7 a b — 4 (a + b), and — 2 a b + 6 (a + b). 22. lsjy-4(a-b), 6 \J y + 2 (a - b), 2 ^ y + (a-b), and sjy — 3(a — b). SUBTRACTION. 19 III. — SUBTRACTION. 61. Subtraction, in Algebra, is the process of finding one of two quantities, when their sum and the other quantity are given. Hence, Subtraction is the converse of Addition. The Minuend is the sum of the quantities. The Subtrahend is the given quantity. The Remainder is the required quantity. As the remainder is the difference between the minuend and subtrahend, subtraction may also be defined as the process of finding the difference between two quantities. 62. Subtraction may be indicated by writing the subtra- hend after the minuend, with a — sign between them. Thus, the subtraction of b from a is indicated by a — b. In indicating subtraction in this way, the subtrahend, if a negative quantity or a polynomial, should be enclosed in a parenthesis. Thus, the subtraction of —b from a is indi- cated by and the subtraction of b — c from a by a— (b — c). 63. Let it be required to subtract b — c from a. According to the definition of Art. 61, we are to find a quantity which when added to b — c will produce a ; this quantity is evidently a — b + c, which is the remainder re- quired. Now, if we had changed the sign of each term of the sub- trahend, giving — b + c, and had added the resulting expres- sion to a, we should have arrived at the same result, a — b + c. 20 ALGEBRA. Hence, to subtract one quantity from another, we may change the sign of each term of the subtrahend, and add the residt to the minuend. 64. 1. Let it be required to subtract 3 a from 8 a. According to Art. 63, the result may be obtained by adding — 3 a to 8 a, giving 5 a (Art. 55). 2. Subtract 8 a from 3 a. By Art. 63, the result is 3 a — S a or —5a (Art. 56). 3. Subtract — 2 a from 3 a. Result, 3 a + 2 a or 5 a. 4. Subtract 3 a from —2 a. Result, — 2a-3«or —5 a. 5. Subtract — 2 a from —5 a. Result, — 5 a + 2 a or —3 a. 6. Subtract —5a from —2 a. Result, — 2 a + 5 a or 3 a. 65. In Arithmetic, addition always implies augmentation, and subtraction diminution. In Algebra this is not always the case ; for example, in adding — 2 a to 5 a the sum is 3 a, which is smaller than 5 a ; also, in subtracting —2 a from 5 a the remainder is 7 a, which is larger than 5 a. Thus, the terms Addition, Subtraction, Sum, and Remainder have a much more general signification in Algebra than in Arith- metic. 66. From Art. 63 we derive the following RULE. To subtract one expression from another, set the subtrahend underneath the minuend, similar terms being in the same ver- tical column. Change the sign of each term of the subtrahend from + to — , or from — to + , and add the restdt to the minuend. USE OF PAKENTHESES. 21 EXAMPLES. 1. 2. 3. 4. 5. 27 a 17 x -13 3, - 10 m n 5 a 2 b 13 a -11b Ay - - 18 m n UaH ab + cd — ax 7x+5y—3a Aab — 3 cd-\- Aax x—7y+ha—A 8. 9. 7 abc-llx + 5y- 48 5 \Ja- 3 y 2 + 7 a -6 llabc+ 3x + 7?/ + 100 3<Ja+ y 2 -5a-7 10. Subtract —5b from - 12 J. 11. From 31 x 2 - 3 y 2 + a b take 17 x 2 + 5 if - 4 a b + 7. 12. Subtract a — b + c from a + b — c. 13. Subtract 6a — 3^ — 5c from Qa-\-3b — 5 c + 1. 14. From 3m-5ft+r-2s take 2 r + 3 n — m — 5 s. 15. Take 4 a — b + 2 i — 5 d from a" — 3 & + a - c. 16. From m 2 + 3 n z take — 4 m' 2 — 6 ?i 8 + 71 cc. 17. From a + b take 2«-25 and — a + b. 18. From a — b — c take — a + b + c and a — b + c. IV. — USE OF PARENTHESES. 67. The use of parentheses is very frequent in Algebra, and it is necessary to have rules for their removal or introduc- tion 22 ALGEBRA. 68. Let it be required to indicate the addition of 3 a. and 5 b — c + 2 d ; this we may do by placing the latter expression in a parenthesis, prefixing a + sign, and writing after the former quantity, thus : 3 a + (5 b - c + 2 d). If the operation be performed, we obtain (Art. 60), 3a + 5b — c + 2d. 69. Again, let it be required to indicate the subtraction of 5b — c-\-2d from 3 a ; this we may do by placing the former expression in a parenthesis, prefixing a — sign, and writing after the latter quantity, thus : 3a-(5b-c + 2d). If the operation be performed, we obtain (Art. 66), 3 a — 5 b + c — 2d. 70. It will be observed that in the former case the signs »f the terms within the parenthesis are unchanged when the parenthesis is removed ; while in the latter case the sign of each term within is changed, from + to — , or from — to +. Hence, we have the following rule for the removal of a paren- thesis : If the parenthesis is preceded by a -f- sign, it may be re- moved if the sign of every enclosed term be unchanged; and if the parenthes-is is preceded by a — sign, it may be removed if the sign of every enclosed term be changed. 71. To enclose any number of terms in a parenthesis, we take the reverse of the preceding rule : Any number of terms may be enclosed in a parenthesis, with a + sign prefixed, if the sign of every term enclosed be un- changed ; and in a parenthesis, with a — sig?i prefixed, if the sign of every term enclosed be changed. USE OF PARENTHESES. 23 72. As the bracket, brace, and vinculum (Art. 25) have the same signification as the parenthesis, the rules for their re- moval or introduction are the same. It should be observed in the case of the vinculum, that the sign apparently prefixed to the first term underneath is in reality the sign of the vin- culum ; thus, + a — b signifies + (a — V), and — a — b signi- fies — (a — b). 73. Parentheses will often be found enclosing others ; in this case they may be removed successively, by the preceding rule ; and it is better to begin by removing the inside pair. 74. 1. Remove the parentheses from 3 a — (2 a — 5) — (-a + 7). Result, 3a — 2a + 5 + a — 7 — 2a — 2. 2. Remove the parentheses etc., from 6 a - [3 a + (2 a - { 5 a - [4 a - a - 2] } )]. In accordance with Art. 73, we remove the vinculum first, and the others in succession. Thus, 6 a — [3 a +(2 a -{5 a -[4 a- a- 2]})] = 6 a - [3 a + (2 a - {5 a - [4 a - a + 2]})] = 6 a — [3 a + (2 a— {5 a — 4 a + a — 2})] = 6 a — [3 a + (2 a — 5 a + 4 a — a + 2)] = Qa — [3a + 2a — 5a + 4a — a + 2^\ = 6a — 3a — 2a + 5a — 4:a + a — 2 = 3a — 2, Ans. 3. Enclose the last three terms of a — b~ c + d+ e —f in a parenthesis with a — sign prefixed. Result, a — b — c—(—d — e+f). 24 ALGEBRA. EXAMPLES. Remove the parentheses, etc., from the following : 4. a — (b — c) + (d — e). 5. 3a-(2a-{a + 2}). 6. 5 x — (2 x — 3 y) — (2 x +. 4 y). 7. a — b + c — {a + b — c) — {c + b — a). 8. in' — 2n+ (a — n + 3 ?m 2 ) — (5 a + 3 w — m 2 ). 9. 2 m - [n — {3 wi -(2w- m) } ]. 10. 8x — (5x — [4 a; — ?/ — &■]) — (— a; — 3 y). 11. 2«-[5J+ {3c-(a+[2ft-3a + 4c])}]. 12. 3c+(2a-[5c-{3a + c-4a}-]). 13. 6 a - [5 a - (4 a - { - 3 a - [2 a- a- 1]})]. 14. 2 m - [3 m - (5 m - 2) - { m - (2 wi -3m + 4)}]. 75. As another application of the rule of Art. 70, we have the following four results : + (+ a) is equivalent to + a ; + ( — a) is equivalent to — a ; — (+ a) is equivalent to — a ; — (—a) is equivalent to + a. V. — MULTIPLICATION. 76. Multiplication, in Algebra, is the process of taking one quantity as many times as there are units in another quantity. The Multiplicand is the quantity to be multiplied or taken. The Multiplier is the quantity by which we multiply. The Product is the result of the operation. The multiplicand and multiplier are often called factors. MULTIPLICATION. 25 77. The product of the factors is the same, in whatever order they are taken. For we know, from Arithmetic, that the product of two numbers is the same, in whatever order they are taken ; thus we have 3 X 4 or 4 X 3 eacli equal to 12. Similarly, in Alge- bra, where the symbols represent numbers, we have a X b or b X a each equal to a b (Art. 14). 78. Let it be required to multiply a — b by c. By Art. 77, multiplying a — b by c is the same as multiply- ing c by a — b. To multiply c by a — &, we multiply it first by a, and then by b, and subtract the second result from the first, e multiplied by a gives a c, and multiplied by b gives b c. Subtracting the second result from the first we have a c — b c the product required. 79. Let it be required to multiply a — b by c — d. To multiply a — b by c — d, we multiply it first by c, and then by d, and subtract the second result from the first. By Art. 78, a — b multiplied by c gives ac — bc, and multiplied by d gives ad — bd. Subtracting the second result from the first, we have ac — bc — ad+bd the product required. 80. We observe in the result of Art. 79, 1. The product of the positive term a by the positive term c gives the positive term a c. 2. The product of the negative term —b by the positive term c gives the negative term —be. 3. The product of the positive terra a by the negative term — d gives the negative term — a d. 4. The product of the negative tern> —b by the negative term — d gives the positive term b d. 26 ALGEBRA. From these considerations we can state what is known as the Rule of Signs in Multiplication, as follows: + multiplied by +, and — multiplied by — , produce + ; + multiplied by — , and — multiplied by + , produce — . Or, as may he enunciated for the sake of hrevity with regard to the product of any two terms, Like signs produce + , and unlike signs produce — . 81. Let it he required to multiply 7 a by 2 b. Since (Art. 77) the factors may he written in any order, we have 7ax2b = 7x2xaXb = 14:ab. Hence, The coefficient of the product is equal to the product of the coefficients of the factors. 82. Let it he required to multiply a 3 hy a 2 . By Art. 17, a s means sX«X« } and a 2 means aXa; hence, a ! Xfl 2 = «X»X«X«X(i=« 5 . Hence, The exponent of a letter in the product is equal to the sum of its exponents in the factors. ' Or, in general, a m X a n = a m + n . 83. In Multiplication we may distinguish three cases. CASE I. • 84. WJien both factors are monomials. From Arts. 80, 81, and 82 is derived the following rule for the product of any two monomials. RULE. Multiply the numerical coefficients together ; annex to the residt the letters of both monomials, giving to curb letter an exponent equal to the sum of its exponents in the factors. Make the product + when the two factors have the same sign, and — when they have different signs. MULTIPLICATION. 27 EXAMPLES. 1. Multiply 2 «" by 3 a 2 . 2« 4 x3a 2 = 6a 6 , Ans. 2. Multiply a 3 b 2 c by - 5 a 2 b d. a 8 b 2 c X — 5 a 2 b d — — 5 a 5 b s c d, Ans. 3. Multiply — 7 x m by — 5 se n . — 7 a; m x — 5 x n = 35 a: m+n , ^4?zs. 4. Multiply 3 a (a; — y) 2 by 4« 3 (x- ?/). 3 a (a - y) 2 X 4 a 3 (x-y) = 12 a 4 (a; - y) 3 , Ans. Multiply the following : 5. 15 m 5 w 6 by 3 m n. 12. — 12 a 2 x by — 2 a 2 y. 6. 3 a 6 by 2 a e. 13. 3 a m 6 n by — 5 a n b r . 7. 17 a b c by — 8 a b c. 14. — 4 x m ?/" by — x n y n z b . 8.-17 a 4 c 2 by - 3 a 2 c 2 . 15. 2 a m 5" by 5 a 3 b. 9. 11 n 2 y by — 5 w 6 «. 16. — 7 m n x 2 by m n # r y 2 . 10. 4a 6 by3aiy 2 . 17. 2 m (a - b) 2 by m (a - b). 11. — 6 a b 2 c by a 3 b m. 18. 7 a (x — y) hy —3 a 2 b (x — y). 19. Find the continued product of 8 a x 2 , 2 a 3 y, and 4 a; 3 v/ 4 . 20. Find the continued product of 2 a c 2 , — 4 a c 3 , and -3«J 2 . CASE II. 85. Wlien one of the factors is a polynomial. From Art. 78 we have the following RULE. Multiply each term of the multiplicand by the multiplier, remembering that like signs produce +, and unlike signs pro- duce — . 28 ALGEBRA. EXAMPLES. 1. Multiply 3 x — y by 2 x y. 3 x — y 2 x y 6 x' 2 y — 2 x y 2 , Ans. 2. Multiply 3 a — 5 x by — 4 ra. 3 a — 5 x — 4 m ,3 .2 — 12 a m + 20 ra #, ^4?is. Multiply tbe following : 3. x 2 -2z-3by 4z. 7. -x 4 -10a; 3 + 5by-2x- 4. 8 a 2 6 c - fZ by 5 a f/ 2 . 8. a 2 + 13 a& - 6 6 2 by 4 a b 2 5. 3 x 2 + 6 x — 7 by — 2 x s . 9. ra 2 + m n + ?r by m n. 6. 3 ra 2 — 5 ra ?i — ?i 2 by — 2 m. 10. 5 — 6 a — 8 a 3 by — 6 <x 11. 5a; 3 -4x 2 -3z-2by-6r\ 12. « 3 - 3 a 2 b + 3 a b 2 - b* by a 2 b 2 . CASE III. 86. When both of the factors are polynomials. In Art. 79 we sbowed that tbe product of a — b and c — d rnigbt be obtained by multiplying a — b by c, and then by d, and subtracting tbe second result from tbe first. It would evidently be equally correct to multiply a — b by c, and then by — d, and add the second result to the first. On this we base the following rule for finding the product of two poly- nomials. RULE. Multiply each term of the multiplicand by each term of the multiplier, remembering that like signs 'produce +, and unlike signs produce —, and add the partial, products. MULTIPLICATION. 29 EXAMPLES. 1. Multiply 3 a - 2 b by 2 a- 5b. 3a -2b 2a —5b 6 a 2 — Aab - 15 a b + 10 b 2 6 a 2 — 19 a b + 10 6' 2 , ^ws. The reason for shifting the second partial product one place to the right, is that in general it enables us to place like terms in the same vertical column, where they are more conveniently added. 2. Multiply x 2 + 1 — x 3 — x by x + 1. 1 — X + X 2 — X 3 1 + x 1 — X + X 2 — X 3 + X — X 2 + X 3 — X* 1 —x 4 , Ans. It is convenient, though not essential, to have both multi- plicand and multiplier arranged in the same order of powers (Art. 41), and to write the product in the same order. Multiply the following : 3. 3 x 2 — 2 x y — y 2 by 2 x — Ay. 4. x 2 + 2x + lhyx 2 -2x + 3. 5. a + b — c by a — b + c. 6. 3a-2bhy-2a + 4.b. 7. a 2 + b 2 + ab by b — a. 8. 1 + x + x 3 + x 2 by a x — a. 9. 5 a 2 - 3 a b + 4 b 2 by 6 a - 5 b. 10. 3 x 2 — 7 x + 4 by 2 x 2 + 9 x — 5. 30 ALGEBRA. 11. 6 x - 2 x- - 5 - a- 3 by x 2 + 10 - 2 x. 12. 2a 3 +5a> 2 -8a:-7by4-5 : * ; -3a: 2 . 13. a 3 b - a 2 b 2 - 4 a 6 8 by 2 a 2 5 - a b 2 . 14. x m + 2 ij — 3xy n ~ l by 4 *"• + 5 y 2 — 4 a 4 y n . 15. 6 a: 4 - 3 x 3 - a: 2 +6 a; -2 by 2x 2 + x + 2. 16. m 4 — m 3 w + rn 2 n 2 — m w 3 + % 4 by m + n. 17. ft 3 -3« 2 H3ai 2 - 6 3 by a 2 -2ab + b\ 87. It is sometimes sufficient to indicate the product of polynomials, by enclosing each of the given factors in a paren- thesis, arid writing them one after the other, with or without the sign X between the parentheses. When the indicated multiplication is performed, the expression is said to be ex- panded or developed. 1. Indicate the product of 2 x 2 — 3 x y + 6 by 3 x 2 + 3 x y — o. Kesult, (2x 2 -3xy + 6) (3x 2 +3xy-5). EXAMPLES. 2. Expand (3 a + 4 b) (2 a + b). 3. Expand (a* — a 3 x + a 2 x 2 — a x 3 + x 4 ) (a + x). 4. Develop (a* — a: 4 ) X (« 4 — a,' 4 ). 5. Develop (a m — a") (2 a — a n ). 6. Expand (1 + x) (1 + a- 4 ) (1 — a- + .r 2 - a- 3 ). 7. Find the value of (« + 2 a) (« — 3 x) (a + 4 x). 8. Expand [« (a 2 — 3 a + 3) - 1] x [a (a - 2) + 1]. 88. From the definition of Art. 76, X a means taken a times. Since taken any number of times produces 0, it follows that x a = 0. That is, If zero be multiplied by any quantity} the product is equal to zero. DIVISION. 31 89. Since (+ a) X (+ b) —ab, and (— a) X (— &) =ab, it follows that in the indicated product of two factors, all the signs of both factors may be changed without altering the value of the expression. Thus, (x — y) {a — b) is equal to {y — x) (J> — a). Similarly we may show that in the indicated product of any number of factors, any even number of factors may In/re their signs changed without altering the value of the expression. Thus, (x — y) (c — d) (e —f) (g — h) is equal to {y-x) (c-d) (f-e) (g- h), or to (y — x) (d — c) {f-e) (h — g), etc.; but is not equal to (y-x){d~c)(f-e)(g-h). VI. — DIVISION. 90. Division, in Algebra, is the process of finding one of two factors, when their product and the other factor are given. Hence, Division is the converse of Multiplication. The Dividend is the product of the two factors. The Divisor is the given factor. The Quotient is the required factor. 91. Since the quotient multiplied by the divisor produces the dividend, it follows, from Art. 80, that if the divisor and quotient have the same sign, the dividend is + ; and if they have different signs, the dividend is — . Hence, + divided by +, and — divided by — , produce + ; + divided by — , and — divided by +, produce — . Hence, in division as in multiplication, Like signs produce +, and unlike signs produce — . 32 ALGEBRA. 92. Let it be required to find the quotient of 14 a b divided by 7 a. Since the quotient is such a quantity as when multiplied by the divisor produces the dividend, the quotient required must be such a quantity as when multiplied by 7 a will produce 14 a b. That quantity is evidently 2 b. Hence, The coefficient of the quotient is equal to the coefficient of the dividend divided by the coefficient of the divisor. 93. Let it be required to find the quotient of a 5 divided by a 3 . The quotient required must be such a quantity as when multiplied by a 3 will produce a 5 . That quantity is evidently a 2 . Hence, The exponent of a letter in the quotient is equal to its expo- nent in the dividend diminished by its exponent in the divisor. Or, in general, a m -f- a n = a m ~ n . 94. If we apply the rule of Art. 93 to finding the quotient of a™ divided by a m , we have a m -f- a m = a m ~ m = a . Now, according to the previously given definition of an ex- ponent (Art. 17), a° has no meaning, and we are therefore at liberty to give to it any definition we please. As a m — a m = 1, we should naturally define a° as being equal to 1 ; and as a may represent any quantity whatever, Any quantity whose exponent is is' equal to 1. By this notation, the trace of a letter which has disappeared in the operation of division may be preserved. Thus, the quotient of a 2 b 3 divided by a 2 b 2 , if important to indicate that a originally entered into the term, may be written a b. 95. In Division we may distinguish three cases. CASE I. 96. When both dividend and divisor are monomials. From the preceding articles is derived the following DIVISION. 33 RULE. Divide the coefficient of the dividend by that of the divisor ; and to the result annex the letters of the dividend, each with an exponent equal to its exponent in the dividend diminished by its exponent in the divisor; omitting all letters whose expo- nents become zero. Make the quotient + when the dividend and divisor have the same sign, and — when they have different signs. EXAMPLES. 1. Divide 9 a 2 b c x y by 3 a b c. 9a 2 bcxy-i-3abc = 3axy, Ans. 2. Divide 24 a 4 m 3 n 2 by — 8 a m 3 n. 24 a 4 m 3 n 2 -. — 8 a m 3 n — — 3 a 3 n, Ans. 3. Divide — 35 x m by — 7 x n . — 35 x m -^- — 7x n = 5 x m ~ n , Ans. Divide tbe following : 4. 12 a 5 by 4 a. 8. - 65 a 3 b 3 c 3 by - 5 a b 2 c 3 . 5. 6 a 2 c by 6 a c. 9. 72 m° n by — 12 m 2 . 6. 14 m 3 n 4 by - 7 m n 3 . 10. - 144 c 5 <P e 6 by 36 c 2 d 3 e. 7. -18x 2 y 5 zhy9x 2 z. 11. - 91 x 4 y 3 z 2 by - 13 x 3 y 1 . CASE II. 97. When the dividend is a polynomial and the divisor is a monomial. Tbe operation being just tbe reverse of that of Art. 85, we have the following RULE. Divide each term of the dividend by the divisor, remembering that like signs produce +, and tinlike signs produce — . 34 ALGEBRA. EXAMPLES. 1. Divide 9 a 3 ft + 6 a 4 e — 12 a ft by 3 a. 3«)9a 3 H6 a 4 c — 12 a 5 3a 2 i + 2« 3 c-4J, ^4»s. Divide the following : 2. 8 a 3 6 c + 16 a 5 6 c — 4 a 2 c 2 by 4 a 2 c. 3. 9 a 5 ft c - 3 a 2 ft + 18 a 3 ft c by 3 a ft. 4. 20 a 4 ft c + 15 a 6 d 3 - 10 a 2 ft by - 5 a ft. 5. 3 a 3 (a - ft) + 9 a (a + ft) by 3 a. 6. 15 (x + y) 2 — 5 a (x + y) + 10 ft (x + y) by — 5 (x + y). 7. 4 x 7 - 8 a- 6 - 14 t> + 2 a 4 - 6 x 3 by 2 x\ 8. 9 a 4 + 27 x 3 - 21 a 2 by -3 a; 2 . 9. _ a 6 £6 C 4 _ a 4 &5 c 3 + 3 a 8 £4 ^2 fcy _ ft 8 p ^ 10. — 12 aP ft? - 30 a 12 ft 3 + 108 a" ft n by - 6 a m ft'". CASE III. 98. When the divisor is a polynomial. 1. Let it be required to divide 12 + 10 x 3 — 11 x — 21 x 2 by 2z 2 -4-3x. We are then to find a quantity which when multiplied by 2 x 2 - 4 - 3 x will produce 12 + 10 x 3 - 11 x - 21 x 2 . Now, in the product of two polynomials, the term containing the highest power of any letter in the multiplicand, multiplied by the term containing the highest power of the same letter in the multiplier, produces the term containing the highest power of that letter in the product. Hence, if the term con- taining the highest power of x in the dividend, 10 x 3 , be di- vided by the term containing the highest power of x in the divisor, 2 x 2 , the result, 5 x, will he the term containing the highest power of x in the quotient. DIVISION. 35 Multiplying the divisor by 5 x, the term of the quotient already found, and subtracting the result, 10 x 3 — 20 x — 15 x 2 , from the dividend,' the remainder, 12 + 9 x — 6 x' 2 , may be re- garded as the product of the divisor by the rest of the quotient. Therefore, to find the rest of the quotient, we proceed as be- fore, regarding 12 + 9 x — 6 x 2 as a new dividend, and divid- ing the term containing the highest power of x, — 6 x' 2 , by the term containing the highest power of x in the divisor, 2 x 2 , giving as a result — 3, which is the term containing the high- est power of x in the rest of the quotient. Multiplying the divisor by — 3, the term of the quotient just found, and subtracting the result, — 6 x 2 + 12 + 9 x, from the second dividend, there is no remainder. Hence, 5 x — 3 is the quotient required. 99. It will be observed that in getting the terms of the quotient, we search for the terms containing the highest power of some letter in the dividend and divisor. These may be obtained most conveniently by arranging both dividend and divisor in order of powers commencing with the highest (Art. 41) ; this, too, facilitates the subsequent subtraction. We also should arrange each remainder or new dividend in the same order. It is customary to arrange the work as follows : 10 x 3 - 21 x 2 - 11 x + 12 10 x s - 15 x 2 - 20 x 2 x 2 — 3 x — 4, Divisor. 5 x — 3, Quotient. — 6x 2 + 9 a: + 12 - 6x 2 + 9 a; + 12 100. We might have obtained the quotient by dividing the term containing the lowest power of x in the dividend, 12, by the term containing the lowest power of x in the divisor, — 4, which would have given as a result — 3, the term containing the lowest power of x in the quotient. In solving the problem in this way, we should first arrange both dividend and divisor in order of powers commencing with the lowest, and should o G ALGEBRA. afterwards bring clown each remainder in the same order; re- membering that a term which does not contain x at all con- tains a lower power of x than any term which contains x. 101. From the preceding principles we derive the follow- ing RULE. Arrange both dividend and divisor in the same order of pow- ers of some common letter. Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient. Multiply the whole divisor by this term, and subtract the product from the dividend, arranging the result in the same order of powers as the divisor and dividend. Regard the remainder as a new dividend, and divide its first term by the first term of the divisor, giving the next term of the quotient. Multiply the whole divisor by this term, and subtract the product from the last remainder. Continue in the same manner until the remainder becomes zero, or until the first term of the remainder will not contain the first term of the divisor. When a remainder is found whose first term will not con- tain the first term of the divisor, the remainder may be written with the divisor under it in the form of a fraction, and added to the quotient. 2. Divide a 8 - 3 a 2 b +. 12 b z + 5 a b 2 by b + a. Arranging the dividend and divisor in order of powers, a + b) a 3 - 3 a 2 b + 5 a b 2 + 12 b z {a 2 - 4 a b + 9 b 2 a 3 + a 2 b - 4 a 2 b ~ — 4 a 2 b — 4 a b 2 9ab 2 9 a b 2 +9b s 3 b 3 , Remainder. 3 b 9 Ans, a 2 -4ab + 9b 2 + a + b' DIVISION. 37 EXAMPLES. 3. Divide 2 a 2 x 2 — 5 a x + 2 by 2 a x — 1. 4. Divide 3 6 3 + 3 a b 2 - 4 a 2 b - 4 « 3 by a + b. 5. Divide 8 a? - 4 a 2 i - 6 a b 2 + 3 & 3 by 2 a - J. 6. Divide 21 a 5 - 21 b 5 by la — lb. 7. Divide a 3 + 2 a; 3 by a + #. 8. Divide x* + y 4 by cc + y. 9. Divide 23 x"- 48 + 6 cc 4 - 2 a; - 31 x* by 6 + 3a; 2 — 5 x. 10. Divide 15 x 4 - 32 x 3 + 50 x 2 - 32 a; + 15 by 3 x 2 - 4 x + 5. 11. Divide 2 a; 4 - 11 aj - 4 ar - 12 - 3 x 3 by 4 + 2 ar + jb. 12. Divide a; 5 — v/ 5 by x — y. 13. Divide 35 - 17 x + 16 x 2 - 25 a; 3 + 6 x 4 by 2 x — 1. 14. Divide 3 x 2 + 4 x + 6 a; - 11 x 3 - 4 by 3 x 2 - 4. 15. Divide a 2 — &' 2 + 2 & c — c 2 by a + b — c. 16. Divide a; 4 — 9 a; 2 — 6 x y — y 2 by x 2 + 3 x + y. 17. Divide x n + 1 + x n y + x y n + y n + 1 by x n + y n . 18. Divide cr n — b 2m + 2 b m c r — c 2r by a n + b m — c r . 19. Divide 1 + a hy 1 — a. In examples of this kind the division does not terminate, there being a remainder however far the operation may be carried. 20. Divide a by 1 + x. 21. Divide a 8 + a 6 b 2 + a 4 b 4 + a 2 b* + b a by a 4 +a 3 b + a 2 b 2 + ab 3 + b 4 . 22. Divide 3 a 3 + 2 - 4 a 5 + 7 a + 2 a 6 - 5 a 4 + 10 a 2 by a 3 — 1 — a 2 — 2 a. 23. Divide 15 x 2 - x 4 - 20 - 2 a; 5 + 6 x + 2 x 3 by 5 - 3 a; 2 - 4 x + 2 x 3 . 38 ALGEBRA. 24. Divide 2 x 5 + 4 x 2 — 14 + 7 x — 7 x 3 + x« — x i by 2 x 2 - 7 + z 3 . 25. Divide 12 « 5 - 14 a 4 b + 10 a 8 i 2 - a 2 6 3 - 8 a i 4 + 4 i 5 by 6 a 8 - 4 a 2 6 - 3 a b 2 + 2 6 3 . 102. In Art. 88 we showed that X a = 0. Since the product of the divisor and quotient equals the dividend, we may regard as the quotient, a as the divisor, and as the dividend. Therefore, °- = 0. That is, a If zero be divided by any quantity the quotient is equal to zero. VII. — FORMULAE. 103. A Formula is an algebraic expression of a general rule. The following formulae will be found very useful in abridg- ing algebraic operations. 104. By Art. 17, (a + bf =5 (a + b) (a + b) ; whence, by actual multiplication, we have That is, ( a + V* = cc 2 + 2ab + b 2 . (1) The square of the sum of two quantities is equal to the square of the first, plus twice the product of the first by the second, plus the square of the second. 105. We may also show, by multiplication, that (a - b) 2 = a 2 -2ab + b\ (2) That is, The square of the difference of two quantities is equal to the square of the first, minus twice the product of the first by the second, plus the square of the second. 106. Again, by multiplication, we have (a + b) (a-b) = a 2 - b 2 . (3) FORMULAE. 3y That is, The product of the sum and difference of two quantities is equal to the difference of their squares. EXAMPLES. 1 107. 1. Square 3 a + 2 b. The square of the first term is 9 a 2 , twice the product of the terms is 12 a b, and the square of the last term is 4 b 2 . Hence, by formula (1), (3 a + 2 b) 2 = 9 a 2 + 12 a b + 4 b 2 , Am. Square the following : 2. 2m + 3 w. 4. 3 x + 11. 6. 2ab + &ac. 3. x 2 + 4. 5. 4a + 5 b. 7. 7 x 3 +3x. 8. Square 4 x — 5. The square of the first term is 16 x 2 , twice the product of the terms is 40 a-, and the square of the last term is 25. Hence, by formula (2), (4 x - 5) 2 = 16 x 2 - 40 x + 25, Am. Square the following : 9. 3a 2 -b s . 11. l-2£ 13. 3-a\ 10. 4 a b - x. 12. x* - y\ 14. 2 a; 3 - 9 x 2 . 15. Multiply 6 a + b by 6 « — 6. The square, of the first term is 36 a 2 , and of the last term b 2 . Hence, by formula (3), (6a + b) (6 a - b) = 36 a 2 - b 2 , Am. Expand the following : 16. (x + 3) (x - 3). 19. (a m + a") (a m - a n ). 17. (2 x + 1) (2 x - 1). 20. O 3 + 5 a-) (a; 3 - 5 a-). 18. (5« + 7J) (5a-76). 21. (4ar + 3) (4a; 2 -3). 22. Multiply a + b + chja + b — c. (a + b + c) (a + b - c) = [(» + b) + c] [O + 6) - c] = (Art. 106), (a + b) 2 - c 2 = a 2 + 2 « b + b 2 - c 2 , Am. 40 ALGEBRA. Expand the following : 23. [1 + («-&)] [1- («- &)]• 25 - (a-5 + c)(a — &-c). 24. (a + & + c) (a — & — c). 26. (c + a - &) (c — a + b). 27. [(a + b) + (c-d)2 [(a + ft) - («-«*)]. 28. (a — b + c — d)(a — b — c + d). 29. (a + b + c + d) (a + b — c — d). VIII. — FACTORING. 108. The Factors of a quantity are such quantities as will divide it without a remainder. 109. Factoring is the process of resolving a quantity into its factors. 110. A Prime Quantity is one that cannot be divided, without a remainder, by any integral quantity, except itself or unity. Thus, a, b, and a + c are prime quantities. Quantities are said to he prime to each other when they have no common integral divisor except unity. 111. One quantity is said to be divisible by another when the latter will divide the former without a remainder. Thus, a b and a b + a 2 b' 1 are both divisible by a, b, and a b. 112. If a quantity can be resolved into two equal factors, it is said to be a, perfect square ; and one of the equal factors is called the square root of the quantity. If a quantity can be resolved into three equal factors, it is said to be a perfect cube ; and one of the equal factors is called the cube root of the quantity. Thus, since 1 a- equals 2 a X 2 a, 4 a 2 is a perfect square and 2 a is its square root ; and since 27 X s equals 3,xx3a , x3/, 27 X s is a perfect cube, and 3x is its <nbe root. FACTORING. 41 Note. 4 a 2 also equals - 2 a x - 2 a, so that the square root of 4 <> 2 may be either 2 a or - 2 «. In the examples in this chapter we shall only consider the positive square root. To find the square root of an algebraic quantity, extract the square root of the numerical coefficient, and divide the exponent of each letter by 2. Thus, the square root of 9 a 6 b 2 is 3 a? b. To find the cube root, extract the cube root of the numerical coefficient, and divide the exponent of each letter by 3. Thus, the cube root of 8 a 3 b® is 2 a b 2 . 113. The factoring of monomials may be performed by- inspection ; thus, 12 a 3 b 2 c = 2.2.8. a a abbe. But in the decomposition of polynomials we are governed by rules which may be derived from the laws of their formation. A polynomial is not always factorable ; but in numerous cases we can always factor ; and these cases, together with the rules for their solution, will be found in the succeeding articles. CASE I. 114. Wlien the terms of a polynomial have a common mo- nomial factor, it may be written as one of the factors of the polynomial, ivith the quotient obtained by dividing the given polynomial by this factor, as the other. 1. Factor the expression 3 x 3 y 2 — 12 x y 4 — 9 x 2 y 3 . We observe that each term contains the factor 3 x y 2 . Dividing the given polynomial by 3 x y 2 , we obtain as a quotient x 2 — 4 y 2 — 3 x y. Hence, 3 x 3 y 2 - 12 x y 4 - 9 x 2 y z = 3 .« y 2 (x 2 -±y 2 -3x y), Arts. EXAMPLES. Factor the following expressions : 2. a s + a. $ 5. 60m 4 n 2 — 12 n\ 3. 16 x 4 - 12 x. 6. 27 c 4 d 2 + 9 c 3 d. 4. a & -2 a 4 + 3 a 3 -a 2 . 7. 36 X s y — 60 x 2 y 4 — 84 x 4 y 2 . 8. a 5 b-3a 6 b 4 -2a 3 b 4 c + 6a' 1 b 5 x. 9. 84 x 2 y 3 - 140 x 3 y 4 + 56 x 4 if. 42 ALGEBRA. 10. 72 o 4 b 3 c 3 + 126 a 3 c 2 d + 162 a 2 c. 1 1. 128 c 4 d 5 + 320 c 2 d 7 - 448 c 8 <Z 4 . CASE II. 115. TFAerc a polynomial consists of four terms, the first two and last two of which have a common binomial factor, it may he written as one of the factors of the polynomial, wit li- the quotient obtained by dividing the given polynomial by this factor, as the other. 1. Factor the expression a m — b m + a • n — b n. Factoring the first two and last two terms by the method of Case I, we obtain m (a — b) + n (a- — b). We observe that the first two and last two terms have the common binomial factor a — b. Dividing the expression by this, we obtain as a quotient m + n. Hence, am — bm + an — bn=(a — b) (m + n), Ans. 2. Factor the expression a m — bm — an + b n. am — b 7)i — a n + b n =■ a m — b m — (a n — b 11) = ni (a — b) — n(a — b) — (a — b) (m — n), Ans. Note. If the third term of the four is negative, as in Ex. 2, it is convenient to enclose the last two terms in a parenthesis with a - sign prefixed, before factoring. EXAMPLES. Factor the following expressions : 3. a b + b x + a y + x y. -7. mx 2 — my 2 — ?ix 2 + n y 1 . 4. a c — cm + a d — dm. 8. x 3 + x 2 + x + 1. 5. x 2 + 2x — xy — 2y. 9. 6 x 3 + 4 x 2 — 9 x - 6. 6. a 3 — a 2 b + a b 2 - b 3 . 10. 8 c x - 12 c y + 2 d x - 3 d y. 11. 6 n - 21 m 2 n-8m + 28 m 3 . 12. a 2 bc — ac 2 d+ab 2 d — bc d 2 . 13. m 2 n 2 x 2 — n s x y — m 3 x y + m n y' 2 . 14. 12 a b m n — 21 a b x y + 20 <■ d m n — 35 c d x //. FACTORING. 4:3 CASE III. 116. When the first and last terms of a trinomial are perfect squares and positive, and the second term is twice the product of their square roots. Comparing with Formulae 1 and 2, Arts. 104 and 105, we observe that such expressions are produced by the product of two equal binomial factors. Reversing the rules of Arts. 104 and 105, we have the following rule for obtaining one of the equal factors : Extract the square roots of the first and last terms, and connect the results by the sign of the second term. 1. Factor a 2 + 2 a b + b 2 . The square root of the first term is a ; of the last term, b ; the sign of the second term is + . Hence, one of the equal factors is a + b. Therefore, a 2 + 2 a b + b 2 = (a + b) (a + b) or (a + b) 2 , Ans. 2. Factor 9 or — 12 a, b + A hi The square root of the first term is 3 a ; of the last term, 2 b ; the sign of the second term is — . Hence, one of the equal factors' is 3 a — 2 b. Therefore, 9 a 2 - 12 a b + 4 b 2 = (3 a -2 b) (3 a - 2 b) or (3 a - 2 bf, Ans. Note. According to Art. 58, the given expression may be written 4 b 2 — 12 a b + 9 a 2 . Applying the rule to this expression, we have 4 b 2 - 12 a b + 9 a 2 = (2 b- 3 a) (2 b - 3 a) or (2 b - 3 a) 2 . We should obtain this second form of the result in another way by apply- ing the principles of Art. 89 to the first factors obtained. EXAMPLES. Factor the following expressions : 3. x 2 - 14 x + 49. 6. a 2 - 28 a + 196.. 4. m 2 + 36 m + 324. 7. n 6 - 26 n 3 + 169. 5. y 2 + 20 y + 100. 8. x 2 y 2 + 32 x y + 256. 44 ALGEBRA. 9. 25 x 2 + 70 x y z + 49 y 2 z 2 . 11. 16 m 2 -8am+« 2 . 10. 36 m? - 36 w « + 9 n 2 . 12. 4 a 2 + 44 a b + 121 b 2 . 13. a 2 6 4 + 12 a b 2 c + 36 c 2 . 14. 9 « 4 + 60 a 2 bc 2 d + 100 b 2 c i d 2 . 15. 4 x A - 60 m re x 2 + 225 m 2 t»s 16. 64 x 6 - 160 x 5 + 100 cc 4 . CASE IV. 117. When an expression is the difference between two perfect squares. Comparing with Formula 3, Art. 106, we observe that such expressions are the product of the sum and difference of two quantities. Reversing the rule of Art. 106, we have the fol- lowing rule for obtaining the factors : Extract the square roots of the first and last terms ; add the results for one factor, and subtract the second result from the first for the other. • 1. Factor 36 x 2 — 49 y 2 . The square root of the first term is 6 a;; of the last, 7 y. The sum of these is 6 x + 7 y, and the second subtracted from the first is 6 x — 7 y. Hence, 36 x 2 — 49 y 2 = (6 x + 7 y) (6 x — 7 y), Ans. 2. Factor {a - b) 2 - (c-df. The square root of the first term is a — b ; of the last, c — d. The sum of these is a — b + c — d, and the second subtracted from the first is a — b — c + d. Hence, (a — b) 2 — (c — d) 2 = (a — b + c — d) (a — b — c + cl), Ans. EXAMPLES. Factor the following expressions : 3. x 2 -l. 5. a 4 -/A 7. 4 a* — 225 m a w 9 . 4. 4x 2 -9t/ 2 . 6. 9 a 2 -4. 8. 1 - 196 x 2 if z\ FACTORING. 45 9. (a + b) 2 - (c + d) 2 . 11. m 2 - (x - y) 2 . 10. (a-c)*—b*. 12. (aj — m) a — (y — »)*. Many polynomials, consisting of four or six terms, may be expressed as the difference between two perfect squares, and hence may be factored by the rule of Case IV. 13. Factor 2 m n + m 2 — 1 + n 2 . Arrange the expression as follows, m 2 + 2mn + r? — 1. Applying the method of .Case III to the first three terms, we may write the expression (m + n)' 2 — 1. The square root of the first term is m + n ; of the last, 1. The sum of these is m + n + 1, and the second subtracted from the first is m + n — 1. Hence, 2 m n + m 2 — 1 + n 2 = (m + n + 1) (m + n — 1), Ans. 14. Factor 2 x y + 1 — x 2 — y 2 . 2 x y + 1 — x 2 — y 2 = l — x 2 -\-2x y — y 2 = 1- {x 2 -2 x y + y 2 ) = l - (x-y) 2 , by Case III, = [1 + (x— y)][l — (as— y)] = (l + #-y) (1— x + y),Ans. 15. Factor 2xy+b 2 -x 2 -2ab- y 2 + a 2 . 2xy+b 2 — x 2 — 2ab — y 2 + a 2 = a 2 — 2ab + b' 2 — x 2 + 2 x y — y 2 = a 2 -2 a b + b 2 - (x s - 2 x y + y 2 ) = (a - b) 2 — (x— y) 2 , by Case III, = [(«-&)+ (as-y)][(a-fl)-(aj-y)] = (a — b+x — y) (a—b — x + y), Ans. Factor the following expressions : 16. x 2 + 2xy + y 2 -4:. 19. 9-x 4 -4?/ 2 + 4f 2 y. 17. a a _ j2 + 2 h c _ c 2. 20. 4 a 2 + 6 2 - 9 d? -4ab. 18. 9 c 2 - 1 + d? + 6 c d. 21. 4 b - 1 - 4 b 2 + 4 m 4 . 22. a 2 — 2 a m + m 2 — b 2 — 2 b n — n 2 . 23. 2 a m — 6 2 + m 2 + 2bn+ a 2 - n 2 . 24. x 2 - y 2 + c 2 - d 2 - 2 c x + 2 tf y. 46 ALGEBRA. CASE V. 118. Wlien an expression is a trinomial, of the form x- + a x + b ; where the coefficient of x 2 is unity, and a and b represent any whole numbers, either positive or negative. To derive a rule for this case we will consider four examples in Multiplication : I. ii. x + 5 x — o x + 3 x — 3 x 1 + 5 x x 2 —ox + 3.x- + 15 -3cc + 15 a; 2 + Sx + 15 sc a -8aT+15 III. IV. x + 5 x — 5 x — 3 x + 3 a? 2 + 5 x # 2 — 5 .<■ -3x-15 +3x-ll « 2 +2^-15 x 2 -2ic-15 We observe in these results, 1. The coefficient of x is the algebraic sum of the numbers in the factors. 2. The last term is the product of the numbers. Hence, in reversing the process, we have the following rule for obtaining the numbers : RULE. Find two numbers ivhose algebraic sum is the coefficient of x, and whose product is the last term. Note. We may shorten the work by considering the following points : 1. When the last term of the product is f, as in Examples I ami II, the sum of the numbers is the coefficient of .<• ; both numbers being + when the second term is +, and - when the second term is -. 2. When the last term is -, as in Examples III and I V, the difference FACTORING. 47 of the numbers (disregarding signs) is the coefficient of x ; the larger number being + and the smaller - when the second term is +, and the larger number - and the smaller + when the second term is - . We may embody these observations in the form of a rule which may be found more convenient than the preceding rule in the solution of examples. I. If the last term is +, find tivo members whose sum is the coefficient of x, and whose product is the last term; and give to both numbers the sign of the second term. II. If the last term is - , find tivo numbers whose difference is the coeffi- cient of x, and whose product is the last term; and give to the larger num- ber the sign of the second term, and to the smaller number tlie opposite sign. 1. Factor x 2 + 14 x + 45. Here we are to find two numbers whose - " - (.product — 45 J The numbers are 9 and 5 ; and, the second term being + , both numbers are +. Hence, x 2 + 14 x + 45 = (x + 9) (x + 5), Ans. 2. Factor x 2 — 6 x + 5. Here we are to find two numbers whose ■! \ (product = 5,' The numbers are 5 and 1 ; and, as the second term is — , both numbers are — . Hence, x 2 — 6 x + 5 = (x — 5) (x — 1), Ans. 3. Factor x 2 + 5 x — 14. Here we are to find two numbers whose ■< " I product = 14 ) The numbers are 7 and 2; and as the second term is +, the larger number is + , and the smaller — . Hence, x 2 + 5 x — 14 = (x + 7) (x — 2), Ans. 4. Factor x 2 — 7 x — 30. Here we are to find two numbers whose ] ~ \ 48 ALGEBKA. The numbers are 10 and 3; and as the second term is — , the larger number is — , and the smaller + . Hence, x 2 - 7 x - 30 = (x - 10) (x + 3), Arts. Note. In case the numbers cannot be readily determined by inspection, the following method will always give them : Eequired two numbers whose difference is 8 and product 48. Taking in order, beginning with the lowest, all possible pairs of integral factors of 48, we have 1x48, 2x24, 3x16, 4x12. And, as 4 and 1 2 differ by 8, they are the numbers required. Evidently this method will give the required numbers eventually, how- ever large they may be, provided they exist. EXAMPLES. Factor the following expressions : 5. a 2 +5 a- + 6. 12. m 2 +9m + 8. 6. a 2 -3a + 2. 13. m' 2 + 2m-80. 7. 2/2 + 2?/- 8. 14. c 2 - 18 c + 32. 8. m 2 -5m-24. 15. x 2 + x-42. 9. * 2 -ll;c + l8. 16. x 2 + 23x + 102. 10. n *- n -c)0. 17. if-9y-90. 11. * 2 +13a; + 36. 18. a 2 +13a-48. 19. cc 2 -9z-70. 20. Factor 15 — 2x — x 2 . 15 _ 2 x - x 2 = - (x 2 + 2 x - 15) By the rule of Case V, x 2 + 2 x - 15 = (.r + 5) (x - 3). Hence, 15 - 2 x - x 2 = - (x + 5) (x - 3) = (x + 5) (3 - x), Ans. Note. If the x* term is -, enclose the whole expression in a paren- thesis with a - sign prefixed. Factor the quantity within the parenthesis, and change the signs of all the terms of one factor. FACTORING. 49 Factor the following expressions : 21. 20-x-x 2 . 22. 6 + x-x 2 . 23. 84-5z-a; 2 . The method of Case V may he extended to the factoring of more complicated trinomials. 24. Factor m 2 ri 2 — 3 m nx + 2 x 2 . r. i it (sum =3) Here we are to find two numbers whose < t — 9 j The numbers are 2 and 1; and as the second term is—, hoth numbers are — . Hence, m 2 ri 2 — 3 m nx + 2x 2 = (m n — 2 x) (m n — x), Ans. Factor the following expressions : 25. a- 4 - 29 x 2 + 120. 30. ?« 4 + 5 m 2 n 2 - 66 n\ 26. c 6 + 12c 3 +ll. 31. (a-b) 2 -S(a-b)-4. 27. x 2 y G + 2xf-120. 32. (x + y)*- 7 (x + y) + 10. 28. (r '^_7«i 2 -144. 33. x 2 - 2 x y 2 z - 48 y* z\ 29. x 2 + 25 w x + 100 ?i 2 . 34. (m + nf + (m + n) - 2. CASE VI. 119. When an expression is the sum or difference of two perfect cubes. By actual division, we may show that a 3 + b 3 a 3 — b 3 — — = a 2 — ab + b' 2 , and — = a 2 + a b + b 2 . a + b a — b Whence, (a 3 + b 3 ) = (a + b) (a 2 -ab + b 2 ), and (a 3 - b 3 ) = (a - b) (a 2 + ab + b 2 ). These results may he enunciated as follows : To factor the sum of two perfect cubes, write for the first factor the sum of the cube 7'oots of the quantities; and for the 50 ALGEBEA. second, the square of the first term of the first factor, minus the product of the two terms, phis the square of the last term. To factor the difference of two perfect cubes, write for the first factor the difference of the cube roots of the quantities ; and for the second, the square of the first term of the first factor, plus the product of the two terms, plus the square of the last term. 1. Factor 8 a 3 + 1. The cube root of the first term is 2 a ; of the last term, 1. Hence, 8 a 3 + 1 = (2 'a + 1) (4 a 2 - 2 a + 1), Ans. 2. Factor 27 x 6 - 64 y 3 . The cube root of the first term is 3 x 2 ; of the last term, 4 y. Hence, 27 x G - 64 y* = (3 x 2 - 4 y) (9 x i + 12 x 2 y + 16 y 2 ), Ans. EXAMPLES. Factor the following expressions : 6. Sc 6 -d 9 . 9. 343 + 8 a 3 . 7. 125 a s - 216 m 3 . 10. 27 a; 3 -125. 8. 729 c 3 dP + 512. 11. 1000 -27 a 3 b\ CASE VII. 120. When an expression is the sum or difference of two like powers ofttvo quantities. ■ The following principles are useful to remember : 1. a n — b n is always divisible by a — b, if n is an. integer. 2. a n — b n is always divisible by a + b, if n is an even integer. 3. a n + b n is always divisible by a + b, if n is an odd integer. We may prove the first principle as follows : Commencing the division of a n — b n by a — b, wv have 3. iC 3 — y 3 . 4. a 3 + 8. 5. m 3 + 64 ?z, 6 . a" — b n FACTORING. 51 a — b a n ' + . . . Quotient. - a n ] b — b n Remainder. a n — b n , a n - } b — b n , b (a n ~ l — b n ~ v ) or, - — - =a n ~ i H ; = a n ~ l -\ '-, a — b a — b a — b It is evident from this result that, if a"" 1 — & n_1 is exactly divisible by a — b, the dividend a n — b n will be exactly divisi- ble by a, — b. That is, if the difference of two like powers of two quantities is divisible by the difference of those quantities, then the difference of the next higher powers of the same quantities is also divisible by the difference of the quantities. But a s — b 3 is divisible by a — b, hence a 4 — b 4 is ; and since c^ — b* is divisible by a — b, a 5 — b 5 is; and so on to any power. This proves the first principle. Similarly the second and third principles may be proved. By continuing the division, we should find, = a"- 1 + a n ~ 2 b + a n ~ 3 b~ + + ab n ~ 2 + b n - 1 (1) a n ~' 2 b + a n - s b 2 — + ab n ~ 2 — fi"" 1 ' (2) a -b 10 a n a — b n + b = a n - -l a n L. + b n — a n ~ -l a + b a 2 b + a n - 3 b 2 - —ab n - 2 +b n ~ 1 (3) It is useful to remember that when a — b is the divisor, all the terms of the quotient are + ; where a + b is the divisor, the terms of the quotient are alternately + and — , the last term being + if n is odd, and — if n is even., 1. Factor a 1 — IP. Putting n — 7 in (1), we have a 1 -b a* + a 5 b + a 4 b 2 + a 3 b 3 + a 2 b* + ab s + b 6 . a — b Hence, a 7 - V = (a - b) (a 6 + a 5 b + a 4 b 2 + a 3 b 3 + a 2 b 4 + a b 5 + b 6 ), Ans. 52 ALGEBRA. 2. Factor m 5 + x 5 . Putting a = m, b = x, n== 5, in (3), we have 1YI -f- X . „ 2 2 3,4 = m — m° x + m x — mx A + x\ m + x Hence, m 5 + x s = (m + x) (ra 4 — m 3 a; + ra 2 a: 2 — m x 3 + x 4 ), Ans. 3. Factor x 6 — y 6 . Putting a = x, b = y, n = 6, in (1), we have x 6 ifi — = x s + x* y + x s y 2 + x 2 y 3 + x y* + y 5 . Hence, x 6 — y 6 = (x — y) (x 5 + x 4 y + x s y 2 + x 2 y s + x y* + f), Ans. Or, putting a =x, b = y, ?i = 6, in (2), we have x 6 ?/ 6 J_ — ^5 ~,i », I ~,3 „fi ™2 „.3 i_ ™ ,.4 „.5 Hence, = x° — a;* ?/ + x" y — x" y 6 + x if — y" x 6 — y 6 =(x + y) (x h — x 4 y + x s y 2 — x 2 y 3 + xy i — y s ), Ans. EXAMPLES. Factor the following expressions : 4. x 5 + y 5 . 6. to 6 -c 6 . 8. m s -n*. 10. a 4 -16. 5. (*-d\ 7. a' + V. 9. c 7 -l. 11. a 7 +128. 121. By the application of one or more of the given rules for factoring, a quantity may sometimes be separated into more than two factors. 1. Factor 2 a x s y 2 - 8 a x y\ By Case I, 2 a x 3 if - 8 a x if = 2 a x y 2 (x 2 - 4 if). Factoring the quantity in the parenthesis by Case IV, 2 a x 3 if — 8 a x if = 2 a x y 2 (x + 2 y) (x — 2 y), Ans. Note. If the method of Case I is to be used in connection with other cases, it should be applied first. GREATEST COMMON DIVISOR. 53 2. Resolve a G — b 6 into four factors. By Case IV, a 6 - b G = (a 3 + b 3 ) (a 3 - b 3 ). By Case VI, a 3 + b 3 = (a + b) (a 2 -ab + b 2 ), and a 3 - b 3 = (a - b) (a 2 + ab + b 2 ). Hence, a 6 -b G =(a + b) (a- b) (a 2 -ab + b 2 ) (a 2 +ab + b 2 ), Ans. EXAMPLES. Factor the following expressions : 3. 3 a 3 b + 12 a 2 b + 12 a b. 7. 3 « 4 - 21 a 3 + 30 a: 4. 45 x 3 if — 120 x 2 f + 80 x if. 8. 2 c 3 m + 8 c 2 ra-42 c m. 5. 18 x 3 y — 2 x f. 9. m 2 a;y-4wa;y— 12 ay. 6. x 3 + Sx 2 +7x. 10. 32 « 4 b + 4 a 6 4 . 11. Resolve n 9 — 1 into three factors. 12. Resolve x i — if into three factors. 13. Resolve x 8 — m 8 into four factors. 14. Resolve m 6 — ?i 6 into four factors. 15. Resolve a 9 + c 9 into three factors. 16. Resolve 64 « 6 — 1 into four factors. Other methods for factoring will he given in Chapter XXIX. IX.— GREATEST COMMON DIVISOR. 122. A Common Divisor or Measure of two or more quan- tities is a quantity that will divide each of them without a remainder. Hence, any factor common to two or more quantities is a common divisor of those quantities. Also, when quantities are prime to each other, they have no common measure greater than unity. 54 ALGEBRA. 123. The Greatest Common Divisor of two or more quan- tities is the greatest quantity that will divide each of them without a remainder. Hence, the greatest common divisor of tivo or more quanti- ties is the product of all the prime factors common to those quantities. By the greatest of two or more algebraic quantities, it may he remarked, is here meant the highest, with reference to the coefficients and exponents of the same letters. In determining the greatest common divisor of algebraic quantities, it is convenient to distinguish three cases. CASE I. 124. When the quantities are monomials. 1. Find the greatest common divisor of 42 a 3 b 2 , 70 a 2 b c, and 98 a 4 b 3 d 2 . 42 a 3 b 2 =2x3x7 a a a bb 70 a 2 be =2x5x7 aa b c 98 a 4 b 3 d 2 = 2 x 7 X 7 aaaabbb del Hence, G. C. D. = 2 X 7 a a b = 14 d 2 b, Ans. (Art. 123). • RULE. Resolve the quantities into their prime factors, and find the product of all the factors common to the several quantities. Note. Any literal factor forming a part of the greatest common divisor will take the lowest exponent with which it occurs in either of the given quantities. EXAMPLES. Find the greatest common divisors of the following : 2. a s x 2 , 7 a* x, and 3 a b 2 . 3. G c 5 d\ 3 c 3 d 5 , and 9 c* d 3 . 4. 18 m n 5 , 45 m 2 n, and 72 m 8 ri 2 . 5. 15 c 2 x, 45 c 3 x 2 , and 60 c 4 x 8 . GREATEST COMMON DIVISOR. 55 6. 108 y 2 z\ 144 f z\ and 120 if z 5 . 7. 96 a 5 b\ 120 a 3 b 5 , and 168 a 4 b & . 8. 51 m 4 n, 85 ra 3 a-, and 119 m 2 if. 9. 84 a 8 y 4 z s , 112 a; 4 v/ 5 z 6 , and 154 a 7 y 6 z\ CASE II. 125. When the quantities are polynomials which can be readily factored by inspection. 1. Find the greatest common divisor of 5xy 3 — 15y 3 , x 2 + 4 x — 21, and mx — 3m — nx + 3n. 5xy 3 — 15y 3 = 5y 3 (x — 3) x 2 + 4 x - 21 = (x + 7) (a; - 3) mx — 3 m — nx + 3n = (m — n) (x — 3) Hence, by Art. 123, G. C. D. = x — 3, Ans. 2. Find the greatest common divisor of 4 x 2 - 4 x + 1, 4 x 2 - 1, and 8 ar 5 - 1. 4a; 2 -4a; + l = (2a;-l) (2 a; - 1) 4 a 2 - 1 = (2 x + 1) (2 x-1) 8 a? - 1 = (2 x - 1) (4 a; 2 + 2 x + 1) Hence, G. C. D. = 2 a: — 1, Ans. The rule in this case is the same as in Case I. EXAMPLES. Find the greatest common divisors of the following : 3. 3 a x 2 — 2 a 2 x, a 2 x 2 — 3 a b x, and 5 a 2 x 3 + 2 ax 4 — 3 a 3 x. 4. vi 2 + 2 in n + n 2 , m 2 — n 2 , and m 3 + n 3 . 5. x* — 1, a- 5 + a 3 , and a; 4 + 2 a; 2 + 1. 6. 3 a xif + 21 ay 2 , 3 c a- + 21 c - 3 d x - 21 rZ, and a; 2 -3a- - 70. 7. 4 x 2 — 12 a + 9, 4 a; 2 - 9, and 4 m 2 »a;-6 m 2 n. 8. 9a 2 -16, 3a;y — 4^ + 3a;2 — 4a, and 27 a; 3 — 64. 56 ALGEBRA. 9. x s — x, X s + 9 x 2 — 10 x, and x 6 — x. 10. a 3 — 8 ft 3 , 5 a x + 2 a — 10 ft x — 4 ft, and a 2 — 4 a ft + 4 ft 2 . 11. ar 2 - a; - 42, x 2 - 4 a - 60, and x 2 + 12 x + 36. 12. 8 x 3 + 125, 4 aj 2 - 25, and 4 a 2 + 20 x + 25. 13. 3 a x G — 3 a x 5 , a x s — 9 a x 2 + 8 a x, and 2 a x h — 2 a x. 14. 12 a x - 3 a + 8 c x - 2 c, 64 x 3 - 1, and 16 as* — 8 a? + 1. CASE III. 126. When the quantities are polynomials which cannot be readily factored by inspection. Let a and ft be two expressions, arranged in order of powers of some common letter ; and let the exponent of the highest power of that letter in ft be either equal to or less than the exponent of the highest power of that letter in a. Suppose that ft is contained in a, p times with a remainder c ; suppose that c is contained in ft, q times with a remainder d ; and suppose that d is contained in c, r times with no remainder. The operation of division may be shown as follows : ft) a (p p ft e) b (q 1 c d) c (r rd We will first show that d is a common divisor of a and ft. From the nature of subtraction, the minuend equals the sub- trahend plus the remainder ; hence, a=pb + c, ft = q c + d, and c = rd. Substituting r d for c in tho value of ft, we have b = q r d + d = d (q r + 1). GREATEST COMMON DIVISOR. 57 Substituting q r d + d for b, and r d for c in the value of a, we have a=p q r d + p d + r d = d (p q r + p + r). Hence, as d is a factor of a and also of b, it is a common divisor of a and b. We will now show that every common divisor of a and b is a divisor of d. Let k he any common divisor of a and b, such that fl^ffli and b = n k. From the nature of subtraction, the minuend minus the subtrahend equals the remainder ; hence, c = a — p b, and d = b — q c. Substituting m k for a, and n k for b in the value of c, we have c = m k — p n k. Substituting mk — pnk for c, and n k for b in the value of d, we have d — nk — q (j)ik—pnk) = nk — qmk+pqnk = k (n — q m + p q n). Hence, k is a factor or divisor of d. Therefore, since every common divisor of a and b is a divisor of d, and no expression greater (Art. 123) than d can be a divisor of d, it follows that d is the greatest common divisor of a and b. 1. Find the greatest common divisor of x 1 — 6 x + 8 and 4 x 3 - 21 x 2 + 15 x + 20. x 2 - 6 x + 8) 4 a 3 - 21 x 2 + 15 x + 20 (4 x + 3 4 x 3 - 24 a; 2 + 32 x 3 x 2 - 17 x + 20 3 a; 2 - 18 a; + 24 x _ 4) ^ — 6x + 8(x-2 x 2 — 4aj . -2a; + 8 -2a + 8 Hence, cc — 4 is the greatest common divisor, Ans. 58 ALGEBRA. RULE. Divide the greater quantity (Art. 123) by the less • and if there is no remainder, the less quantity trill he the required greatest common divisor. If there is a remainder, divide the divisor by it, and continue thus to make the preceding divisor the dividend, and the re- mainder the divisor, until a divisor is obtained which leaves no remainder ; the last divisor will be the greatest common divisor required. Nota 1. If there are three or more quantities, find the greatest common divisor of two of them ; then of this result and the third of the quantities, and so on. The last divisor will be the greatest common divisor required. Note 2. If a monomial factor is seen by inspection to be common to all the terms of one of the given quantities, and not of the other, it may be re- moved, as it evidently can form no part of the greatest common divisor ; and, similarly, we may remove from a remainder any monomial factor which is not a common factor of the given quantities. 2. Find the greatest common divisor of 6 a x~ — 19 a x + 10 a and 6 x s — x 2 — 35 x. In the first quantity a is a common factor of all the terms, and is not a factor of the second quantity ; in the second quan- tity x is a common factor of all the terms, and is not a factor of the first quantity. Hence we may remove a from each term of the first quantity, and x from each term of the second. 6 a; 2 - 19 a + 10)6 cc 2 - x -35(1 6a: 2 -- 19:c + 10 18a;-45 In this remainder 9 is a common factor of all the terms, and is not a common factor of the given quantities. Hence 9 may be removed from each term of the remainder. 2 x - 5)6 x 1 - 19 x + 10(3 x - 2 6 x 2 — 15 x — 4 x'+ 10 — 4 a; + 10 Hence, 2 x — 5 is the greatest common divisor, Ans. GREATEST COMMON DIVISOR. 59 Note 3. If the first term of a remainder be negative, the sign of each term may be changed. 3. Find the greatest common divisor of 2 x 2 — 3 a; — 2 and 2a; 2 -5a;-3. 2a; 2 - 3 x - 2)2 a 2 - 5 x -3(1 2x 2 -3x-2 -2x-l The first term of this remainder being negative, we change the sign of each term, giving 2 x + 1. 2 x + 1)2 a;' 2 - 3 x - 2 (x — 2 2 x' 2 + x — 4x — 2 — 4a; — 2 Hence, 2 x + 1 is the greatest common divisor, Ans. Note 4. The dividend or any remainder may be multiplied by any quantity which is not a common factor of all the terms of the divisor. 4. Find the greatest common divisor of 2 a- 3 — 7 xr + 5 x — 6 and 3 a' 3 — 7 a' 2 — 7 x + 3. To avoid a fraction as the first term of the quotient, we multiply each term of the second quantity by 2, giving 6 X s - 14 x 2 - 14 x + 6. 2 x s — 7 x 2 + 5 x — 6)6 x s - 14 cc 2 - 14 x + 6 (3 6 a; 3 - 21 x 2 + 15 a; - 18 7 a- 2 -29 a; + 24 To avoid a fraction as the first term of the next quotient, we multiply each term of the new dividend by 7, giving 14 x 3 - 49 x 2 + 35 x - 42. 7 a; 2 - 29 x + 24) 14 x 3 - 49 x 2 + 35 x - 42 (2 a; 14 a; 3 - 58 a 2 + 48 x 9 a- 2 - 13 x - 42 60 ALGEBRA. The first term of this remainder not heing exactly divisible by the first term of the divisor, we multiply each term hy 7, giving 63 x 2 — 91 x — 294. 7 x 2 - 29 x + 24) 63 x 2 - 91^-294(9 63 x 2 - 261 x + 216 170 x - 510 Dividing each term by 170, x — 3) 7 x 2 — 29 x + 24 (7 x — 8 7 x 2 -21x — 8 a; + 24 - 8cc + 24 Hence, x — 3 is the greatest common divisor, Ans. Note 5. When the two given quantities have a common monomial factor, it may be removed from each, and the greatest common divisor of the resulting expressions found. This result must be multiplied by the common monomial factor to give the greatest common divisor of the given quantities. 5. Find the greatest common divisor of 6 x 3 — x 2 — 5 x and 21 x 3 - 26 x 2 + 5x. The quantities have the common monomial factor x ; remov- ing it, we find the greatest common divisor of 6 x 2 — x — 5 and 21 x 2 — 26 x + 5. We multiply the latter by 2, to avoid a frac- tion as the first term of the quotient, giving 42 a; 2 — 52 x + 10. 6 x 2 - x - 5) 42 x 2 - 52 x + 10 ( 7 42 x 2 - 7a;-35 — 45 x + 45 Dividing by — 45, x — 1)6 x 2 — x — 5(6a: + 5 6 x 2 — 6 x 5x — 5 5x — 5 LEAST COMMON MULTIPLE. 61 Hence, x — 1 is the greatest common divisor of 6 x 2 — x — 5 and 21 x 2 — 26 x + 5. Multiplying by x, the common mo- nomial factor, we obtain x (x — 1) or x 2 — x as the required greatest common divisor, Ans. EXAMPLES. Find the greatest common divisors of the following : 6. 6 x 2 — 1 x — 24 and 12 x 2 + 8 x - 15. 7. 24 x 2 + 11 x - 28" and 40 x 2 - 51 x + 14. 8. 2 x s -2 x 2 - 3 x + 3 and 2 x 3 -2 x 2 - ox + 5. 9. 6 x 2 - 13 x -28 and 15 x 2 + 23 a; + 4. 10. 8 x 2 - 22 x + 5 and 6 a; 2 - 23 x + 20. 11. 5 x 2 + 58 jc + 33 and 10 .x 2 + 41 x + 21. 12. x 3 + 2 x 2 + x + 2 and x i -4r-x-2. 13. 2a; 3 -3cc 2 -;c + l and 6x s — x 2 + 3x -2. 14. a,- 4 - x 3 + 2 x 2 + x + 3 and x i + 2 x 3 — x — 2. 15. ft 2 -5ax + 4 x 2 and a 3 — « 2 x + 3 a x 2 — 3 x 3 . 16. x A -x 3 -5 x 2 + 2 a- + 6 and x 4 + x 3 - x 2 -2x- 2. 17. 6 x 2 y + 4 x y 2 — 2 y 3 and 4 cc 3 + 2 x 2 ?/ — 2 a* //\ 18. 2« 4 + 3a 3 x-9a 2 x 2 and 6 « 3 - 17 a 2 ;c + 14 a ar- 3x 3 . 19. 15 a 2 x 3 - 20 a 2 x 2 - G5 a 2 x - 30 a 2 and 12 6 a; 3 + 20 b x 2 — 16 b x — 16 b. X. — LEAST COMMON MULTIPLE. 127. A Multiple of a quantity is any quantity that can be divided by it without a remainder. Hence, a multiple of a quantity must contain all the prime factors of that quantity. 62 ALGEBRA. 128. A Common Multiple of two or more quantities is one that can be divided by each of them without a remainder. Hence, a common multiple of two or more quantities must contain all the prime factors of each of the quantities. 129. The Least Common Multiple of two or more quanti- ties is the least quantity that can be divided by each of them without a remainder. Hence, the least common multiple of two or more quantities must be the pmxluct of all their different prime factors, each taken only the greatest number of times it is found in any one of those quantities. By the least quantity, is here meant the lowest with refer- ence to the exponents and coefficients of the same letters. In determining the least common multiple of algebraic quantities, we may distinguish three cases. CASE I. 130. When the quantities are monomials. 1. Find the least common multiple of 36 a s x, 60 a 2 y 2 , and 84 c x s . 36 a 3 # = 2x2x3x3 a a ax 60aV=2x2x3x5 a a yy 84 c x z =2x2x3x7 x x x c Hence, L. C. M. = 2 x f 2 X3x3x5 X? aaaxxxyyc = 1260 a 3 x s if c, Ans. (Art. 129). RULE. Resolve the quantities into their prime factors; and the product of these, taking each factor only the greatest number of times it enters into any one of the quantities, will be the least common multiple. Any literal factor forming a part of the least common mul- tiple will take the highest exponent with which it occurs in p.ither of the given quantities. LEAST COMMON MULTIPLE. 6 o When quantities are prime to each other, their product is their least common multiple. EXAMPLES. Find the least common multiples of the following : 2. 8 a 4 c, 10 a 3 b, and 12 a 2 b 2 . 3. 5 x 3 y, 10 if z, and 15 x z 3 . 4. a 5 b 2 , 9 a 3 b\ and 12 a 2 b 3 . 5. 24 m 3 x 2 , 30 n 2 y, and 32 x y 2 . 6. 8 c 2 d 3 , 10 a e, and 42 a 2 d. 7. 36 x y 2 z 3 , 63 x 3 y z 2 , and 28 .r 2 y 3 z. 8. 40 a 2 b d 3 , 18 a c 3 d\ and 54 J 2 c d\ 9. 7 m w 2 , 8 x 3 y 2 , and 84 n x y 3 . CASE II. 131. When the quantities are polynomials which can be readily factored by inspection. 1. Find the least common multiple of x 2 + x — 6, x 2 — 6 x + 8 and x 2 — 9. x * + a, _ 6 = (x - 2) (x + 3) a;2_6a: + 8 = (a;-2) (cc-4) x 2 -9 =(x-3)(x + 3) Hence (Art. 129), L. C. M. = (x - 2) (a; - 3) (sc + 3) (x - 4) or, x 4 - 6 x 3 - x 2 + 54 x - 72, ^ws. The rule is the same as in Case I. EXAMPLES. Find the least common multiples of the following : 2. a x 2 + a 2 x, x 2 — a 2 , and x 3 — a 3 . 3. 2 a 2 + 2 a b , 3 a b - 3 & 2 , and 4 a 2 c — 4 ft 2 c. 64 ALGEBRA. 4. x 2 + x, x z — x, and x i + x. 5. 2 - 2 a- 2 , 4 - 4 a-, 8 + 8 a, and 12 + 12 a 2 . 6. x 2 + 5x + 4, x 2 + 2x — S, and a; 2 + 7 jc + 12. 7. x 3 — 10 a; 2 + 21 a;, and a x 2 + 5 a x — 24 a. 8. 4 ar - 4 a; + 1, 4 x 2 - 1, and 8 a 3 - 1. 9. a x — a y — b x + b y, x 2 — 2 x y + y 2 , and 3 arb — 3ab 2 . 10. 9 a; 2 + 12 a; + 4, 27 x 3 + 8, and 6 a x 3 + 4 a x 2 . 11. cc 2 - 4 cc + 3, x 2 + a; - 12, and x 2 -x- 20. 12. x 2 — y 2 — z 2 + 2 y z and x 2 -i/ 2 +r + 2a;s. CASE III. 132. When the quantities are polynomials which cannot be readily factored by inspection. Let a and b be two expressions ; let d be their greatest com- mon divisor, and m their least common multiple. Suppose that d is contained in a, x times, and in b, y times ; then, from the nature of the greatest common divisor, x and y are prime to each other. Since the dividend is the product of the quo- tient and divisor, we have a = dx and b = d y. Then (Art. 129) the least common multiple of a and b is d x y, or m = d x y ; but dx = a, and y = -', substituting, we h have m = a X -; • d In a similar manner we could show that m — b X -;• d Hence the following RULE. Find the greatest common divisor of the two quantities ; di- vide one of the quantities by this, and multiply the quotient by the other quantity. LEAST COMMON MULTIPLE. 65 Note. If there are three or more quantities, find the least common multiple of two of them, and then of that result and the third quantity ; and so on. 1. Find the least common multiple of 6 x 2 — 17 x + 12 and 12 a: 2 - 4 a -21. 6 a; 2 - 17 x + 12)12 a: 2 - 4sc-21(2 12 x 2 - 34 x + 24 30 x - 45 2 x - 3 ) 6 x 2 - 1 7 x + 1 2 ( 3 x - 4 6 x 2 — 9 x - 8 x + 12 - 8 a + 12 Hence, 2 a: — 3 is the greatest common divisor of the two quantities ; dividing the first given quantity by this, we obtain, as a quotient, 3 x — 4 ; multiplying the second given quantity by this quotient, we have (3 a; -4) (12 a; 2 -4 a: -21), or 36 x 3 - 60 x 2 - 47 x + 84 as the required least common multiple, Ans. EXAMPLES. Find the least common multiples of the following : 2. 6 x 2 + 13 x - 28 and 12 x 2 - 31 x + 20. 3. 8 x 2 + 30 x + 7 and 12 x 2 - 29 x - 8. 4. a 3 + a 2_ 8 a _ 6 and 2 a 3 - 5 a 2 - 2 a + 2. 5. 2 x 3 + x 2 - x + 3 and 2 .t 3 + 5 x 2 - x - 6. 6. (t 3 -2« 2 H2ffii 2 - 6 3 and a 3 + a 2 6 - a b 2 - b 3 . 7. x* + 2 x 3 + 2 x 2 + x and a « 3 — 2 a x — a. 8. 2x 4 -llx- 3 +3a; 2 + 10a; and 3a; 4 - 14z 3 - 6ar+ 5oj. 66 ALGEBRA. XL — FRACTIONS. 133. A Fraction is an expression indicating a certain number of the equal parts into which a unit has been divided. The denominator of a fraction shows into how many parts the unit has been divided, and the numerator how many parts are taken. 134. A fraction is expressed by writing the numerator above, and the denominator below, a horizontal line. Thus, - is a fraction, signifying that the unit has been divided into b equal parts, and that a parts are taken. The numerator and denominator are called the terms of a fraction. Every integer may be considered as a fraction whose denomi- a nator is unity ; thus, a = r- . 135. An Entire Quantity is one which has no fractional part ; as, ab, or a — b. 136. A Mixed Quantity is one having both entire and b a fractional parts ; as, a , or c + x + y 137. If the numerator of a fraction be multiplied, or the de nominator divided, by any quantity, the fraction is multi- plied by that quantity. 1. Let y denote any fraction ; multiplying its numerator by c, we have -— . Now, in - and — the unit is divided into b b b b equal parts, and a and a c parts, respectively, arc taken. Since FRACTIONS. 67 c times as many parts are taken in — as in - , it follows that a c . .,. a — - is c times -. b o 2. Let — denote any fraction ; dividing its denominator a a a by c, we have -. Now, in — and -, the same number of b be a b parts is taken ; but, since in — tbe unit is divided into a . hc b c equal parts, and in - into only b equal parts, it follows that • a . . , i , • a tt a each part m - is c times as large as each part m — . Hence, - is c times -r— . be 138. If the numerator of a fraction be divided, or the de- nominator multiplied, by any quantity, the fraction is divided by that quantity. 1. Let — denote any fraction ; dividing its numerator by c, we have-. Now, in Art. 137, 1, we showed that — was c a o a . ac . . b times -. Hence, - is — divided by c. b b . b 2. Let - denote any fraction : multiplying its denominator b a a by c, we have — . Now, in Art. 137, 2, we showed that - was a bc a a . . h e times — . Hence, — is - divided by c. b c b c b 139. If the terms of a fraction be both multiplied, or both divided by the same quantity, the value of the fraction is not altered. For, multiplying the numerator by any quantity, multiplies the fraction by that quantity ; and multiplying the denomi- nator by the same quantity, divides the fraction by that quantity. And, by Art. 44, Ax. 6, if any quantity be both multiplied and divided by the same quantity, its value is not altered. 68 ALGEBRA. Similarly, we may show that if Loth terms are divided by the same quantity, the value of the fraction is not altered. 140. We may now show the propriety of the use of the fractional form to indicate division, as explained in Art. 16 ; ft that is, we shall show that - represents the quotient of a di- vided by b. For, let x denote the quotient of a divided by b. Then, since the quotient, multiplied by the divisor, gives tbe dividend, we have b x = a. But, by Art. 137, bXj=a. Therefore, x = - . b 141. A fraction is positive when its numerator and de- nominator have the same sign, and negative when they have different signs. For, a fraction represents the quotient of its numerator divided by its denominator ; consequently its proper sign can be determined as in division (Art. 91). 142. The Sign of a fraction is the sign prefixed to its dividing line, and indicates whether the fraction is to be added or subtracted. Thus, in x -\ — — the sign + denotes that the fraction -j— , although essentially negative (Art. 91), is to be added to x. The sign written before the dividing line of a fraction is termed the apparent sign of the fraction ; and that de] tending upon the value of the fraction itself is termed the real sign. Thus, in -\ — — , the apparent sign is + , but the real sign is — . Where no signs are prefixed, plus is understood. ah — ah ah -ah h b -b -b ah a b — ah — ah — _ i b -b b -b FRACTIONS. 69 143. From the principles of Arts. 140 and 141 we obtain, + a; — a. — o From which it appears that, Of the three signs prefixed to the numerator, denominator, and dividing line of a fraction, any two may he changed with- out altering the value of the fraction ; hut if any one, or all three are changed, the value of the fraction is changed from + to — , or from — to + . 144. If either the numerator or denominator of the frac- tion is a polynomial, we mean by its sign the sign of the entire expression, as distinguished from the sign of any one of its individual terms • and care must be taken, pn changing signs in any such case, to change the sign before each term. „,, a — h—a + b b — a Thus ' -c^r-c^d> ov c—}v a—h—a+b h — a also, j = -., or . c — d —c + d d — c 145. From Art. 141 we have abed _(-a)h (- c) (—d) __ a (- b) (- c) d efffh (~e)fgh e(-f)g(-h) , etc. ; abed _ (— a) bc{— d) _ a (— h) (— c) d ~ Jff h - ( r e)fgY ~ e (-/) (- g) (- h)' : From which it appears that, If the terms of a fraction are composed of any number of factors, any even number of factors may have their signs changed without altering the value of the fraction • but if any 70 ALGEBRA. odd number of factors have their signs changed, the value of the fraction is changed from + to — , or from — to +. a — b a — b b — a Thus, (x — y)(x — z)~ (y — x) (z — x)~ {y^- x) {x — z) b — a 1 n . b — a but does not equal (x — y)(z — x) (y — x){z — x) REDUCTION OF FRACTIONS. 146. Reduction of Fractions is the process of changing their forms without altering their values. TO REDUCE A FRACTION TO ITS SIMPLEST FORM. 147. A fraction is in its simplest form, when its terms are prime to each other. CASE I. 148. When the numerator and denominator can be readily factored by inspection. Since dividing hoth numerator and denominator by the same quantity, or cancelling equal factors in each, does not alter the value of the fraction (Art. 139), we have the fol- lowing RULE. Resolve both numerator and denominator into their prime factors, and cancel all that are common to both. EXAMPLES. , ^, , 18 a 8 b 2 o . . , , 1. ixeduce — — ^— 2 — ™ ^ s simplest form. 18 a*b 2 c __ 2 . 3 . 3 . a . a . a . b . h . c 2ac 45 a 2 b' 2 x ~ 5 . 3 . 3 . a . a . b . b . x 5 x ' * x 2 + 2 x — 15 2. Eeduce — s — — to its simplest form. x- — 2x — 3 g; 2 + 2x -15 _ (x + 5) (x - 3) _ ,-r + 5 x 2 - 2 x - 3 : " (x + 1) (x - 3) ~ x~+l' FRACTIONS. 71 _ _. , b c, — a r — b d + a d . . . 3. Eeduce : ; — to its simplest form. a in — b in — an + n be — ac — bd + ad (b — a) (c — d) am — b m — an+ bn (a — b) (m — n) = (Art. 89) ("-f)(<*-«0 = fLll t Ans . (it — 0) (m — n) m — n Note. If all the factors of the numerator be removed by cancellation, the number 1 (being a factor of all algebraic expressions) remains to form a numerator. If all the factors of the denominator be removed, the result will be au entire quantity ; this being a case of exact division. Reduce the following to their simplest forms : 4. SfUU. 13. c K.9U !IV IV - - oo mr n 6 65x 2 y 3 z* 2(> x* y° z~ „ 54 a 3 b 5 c 2 72 a 2 b 2 c Wmxif ,« o. — — • 1 1 . to m x y- 110 e 3 x 2 y 9 - 22c 2 x 2 ■ 1S ' 1A 2a 2 cd+2abcd 1Q 1U - a~^ Tr — 7 ' iy> b <r x y + b ab,x y ii. 3»'-e« 4 y go. 6x 2 y 2 — 12 xy 3 ac + ad — b c — b d 19 x 2 — 2x — lh' 2mx + 3my — 2n 2 x — 3n 2 y x 2 + 10 x + 21 ' 2 m 2 x + 3 m 2 y—2nx—3n y m 2 — 10m + 16 m 2 + m - -72 * 4 c 2 -20c ' + 25 25-4 c 2 ' 4 a — 9 a n 2 9bn 2 -12bn + 4:b 8 x 3 + y 3 4 x 2 — y 2 ' 27 y 3 - 125 25- -30 7/ + 9y 2 ' 6 a; 2 y — 2 x 3 y x 2 - -8x + 15 " 4 — x 2 X 3 - - 9 x 2 + 14 x ' a e — b c — ad + bd 72 ALGEBRA. CASE II. 149. When the numerator and denominator cannot be readily factored by inspection. Since the greatest common divisor of two quantities con- tains all the prime factors common to both, we have the fol- lowing KULE. Divide both numerator and denominator by their greatest common divisor. EXAMPLES. 1. Reduce — - — 5 — — to its simplest form. 6 cr — a — 12 By the rule of Art. 126, we find the greatest common divisor of the numerator and denominator to he 2.a — 3. Dividing the numerator by this, the quotient is a — 1. Dividing the denominator, the quotient is 3 a + 4. Therefore, the simplest form of the fraction is -, Ans. 3 a + 4' Reduce the following to their simplest forms : 6a: 2 + a:-35 „ 6 a- 3 - 19 x 2 + 7 x + 12 o -«-" " — "- — "J- g ' '2u 2 -7a + 6' . 2 m 2 — 5 m + 3 Q ' 12 m 3 - 28 m + 15 ' x a + x *-Sx-2 O. -= ; s ~ 7Z • 1U. 6 ' 2x 3 + 5x 2 -2:c + 3" 8x 2 + 22 a: + 5' 10 a 2 - -a-21 2u 2 - 7 a + 6' 2 m 2 — 5 m + 3 12 m 2 - - 28 m + 15 ' x s + x 2 — 3 x - -2 x 3 — 4 x 2 + 2x- 4-3" 6 a 3 — 7 x 2 + 5 a •-2 6a; 3 -25a: 2 + 17a- + 20 4a: 3 + 14ar+12a- + 5 4a; 3 -10a; 2 -12a'- rr' 1 12 a 2 + 16 a - 3 10 a 2 + a - 21 ' x 3 — 4 x 2 + 4 a; — 1 a 3 - 2 x 2 + 4 aj - 3 " 6 x 3 — x 2 — 7 x — 2 FRACTIONS. 73 TO REDUCE A FRACTION TO AN ENTIRE OR MIXED QUANTITY. 150. Since a fraction is an expression of division (Art. 140), we have the following RULE. * Divide the numerator by the denominator, and the quotient will be the entire or mixed quantity required. EXAMPLES. ax — a 2 x 2 1. Reduce — to an entire quantity. (ax — a 2 x 2 ) -i-ax = l — ax, Ans. q% A3 /j.3 2. Reduce to a mixed quantity. b 3 a — x)a 3 — x 3 — b 3 (a 2 + a x + x 2 , Ans. a — x a 3 — a 2 x a 2 x — x 3 — b 3 9 a x — a x 2 ax 2 — X 3 — b z ax 2 — x 3 -b 3 Reduce the following to entire or mixed quantities : ab — a 2 o~ 2 ■ k o. 4. 5. 6. 7. 2ab b ' X 3 + y 3 X + y 2: K 2 -3x-4: 5 x X 3 2 i rr — x' + 7x — 6 3 x a 2 — 3ab + 4b 2 8. x — 3 9. X s — 1 X-V 10. 4,x 2 -2x + 5 2x 2 -x + \ ' 11 /Y»<* ___ ryi"— /y» i i _ */ *As *Aj *as *J X 2 + X — 1 12 2cc 3 -3.x 2 +4r-2 2 x 2 — 3 x + 3 74 ALGEBRA. TO REDUCE A MIXED QUANTITY TO A FRACTIONAL FORM. 151. This is the converse of Art. 150; hence we may- proceed by the following RULE. Multiply the entire part by the denominator of the fraction ; add the numerator to the product when the sign of the fraction is + , and subtract it when the sign is — / writing the result over the denominator. EXAMPLES. a 2 £2 5 1. Eecluce a + b — : — to a fractional form. a — b By the rule, a 2 _ b 2 _ 5 ( a + &) (g-b)- Q 2 - b°- - 5) a — b a — b a — b a — b Note. It will be found convenient to enclose the numerator in a pa- renthesis, when the sign before the fraction is — . Reduce the following to fractional forms: 4 „ 3a 2 -2Z> 2 2. x + l + .- 7.2a ^- x — 6 a a 3. a + — 8. a 2 +ab + b 2 — 7 • n b — a 4. 7a; - 4 " 2 + 5a - 9. 3z-2- 3 ~8 2x-l i x + 1 in / a * + b * 5. * + i + __. 10. a -&_-^-. a +■ « a; — ^s FRACTIONS. 75 TO REDUCE FRACTIONS TO A COMMON DENOMINATOR. t co 1 x> i 5cd 3mx j 3 n y 152. 1. Keduce - — — , , and — -^ to a common 3 crb 2 ab 2 ka 6 b denominator. Since multiplying each term of a fraction by the same quan- tity does not alter the value of the fraction (Art. 139), we may multiply each term of the first fraction by 4 a b, giving 20 a b c d , « , 1 ■, -. „ . . 18 a 2 m x ; each term of the second by b a , giving ; 12 a? 6 2 ' J ' & ° 12« :i 6 2 and each term of the third by 3 b, giving L. . 12 a J b" It will be observed that the common denominator is the least common multiple of the given denominators, which is also called the least common denominator ; and that each term of either fraction is multiplied by a quantity which is obtained by dividing the least common denominator by its own denomi- nator. Kence the following RULE. Find the least common multiple of the given denominators. Divide this by each denominator, separately, and multiply the corresponding numerators by the quotients ; writing the results over the common denominator. Before applying the rule, each fraction should be in its sim- plest form ; entire and mixed quantities should be changed to a fractional form (Arts. 134 and 151). Note. The common denominator may be any common multiple of the given denominators. The product of all the denominators is evidently s common multiple, and the rule is sometimes given as follows : "Multiply 'each numerator by all the denominators except its own, and write the results over the product of all the denominators." an a x x it 2. Reduce - — — , — — , and . . ■ N , to a common de- 1 — x (1 — x) 2 (1 — xy nominator. 76 ALGEBRA. The least common multiple of the given denominators is (1 — be) 3 . Dividing this by the first denominator, the quotient is (1 — a:) 2 ; dividing it by the second denominator, the quo- tient is (1 — x) ; and dividing it by the third denominator, the quotient is 1. Multiplying the corresponding numerators by these quotients, we obtain a y (1 — x) 2 , a ar (1 — x), and x i/ 3 as the new numerators. Hence the results are a y (1 — a?) 2 ax 2 (1 — x) x y 3 (i-xy > {i-xy > and (i - xy ' Ans - EXAMPLES. Reduce the following fractions to a common denominator : a 3ab 2ac -,56c _ 4c — 1 3b — 2 1 5a 3 - -^o— > -n— > and "To-- 6 - o „ , > K „ . , and 8 ' 9 ' 12 ' 3ab ' 5ac ' 6b c . x 2 y xy z 7 y z 2 „ 2 3 4 ^** ~t , * i t~^ i and ^"tt — . /. — - — — , — -, and — -. 10 ' 15 ' 30 a 3 a; 2 ' a a: 3 ' at a 2 x 3y z Axz 5a;?/ 5 az 3bx ley — m b ' Yx-'Yy-' and TT " 8 - 6^' 87i' and 10* s 2 9. -, — — -, and — a — b ' a + b ' a 2 + 6 2 ° 10 # + 3 a? + 1 a; + 2 a; 2 - 3 a: + 2' a; 2 - 5x + 6' x 2 - 4a; + 3' 2a 3b 4c cr + a — 6 ' a 2 + 5a + 6' an< " a 2 — 4' 12. T , — — T , and -j — T . a; — 1 ar— 1 x 3 — 1 -n a & m — n a + b a m — b m + a n — b n' 2 a 2 — 2 a b ' 3 b m + 3 b n 14. Reduce ; r^— - , — ;— - , and (a-b) (a-c) ' (b — a) (b-e)' (c-a)(c-b) to u common denominator. FRACTIONS. 77 The fractions may be written (Art 145) as follows : , and (a - 6) (a - c) ' (a -b)(b- e) ' (a - c) (6 - c) The least common denominator is now (a — b) (a — e) (b — c). Applying the rule, we have the results, Q-a)(b-c) (h-l)(a-c) ^ ^ (a -b) (a- c) (b — c)' (a- b) (a - c) (b - c) (l-c)(a-b) (a -b){a- c) (b - c) Kecluce to a common denominator : ,_ 3 2 .a — 2 3 , Ans. a — V a + V 1 2- X 1 + X c > x — + d 1' 16. z , 7, and x' 1 — x . b — a 17. -. j^-. pr-, -p. sT? ~ , and (a + b)(a-b)' (b-a)(c-d)' (d-c)(a + b) 153. A fraction may he reduced to an equivalent one hav- ing a given denominator, by dividing the given denominator by the denominator of the fraction, and multiplying both terms by the quotient. 1. Eeduce — ry to an equivalent fraction having ar — a b + b" a 3 + b 3 for its denominator. (a* + b 3 ) -j- (a 2 -ab + b 2 ) = a + b; multiplying both terms by a + b, a — b (a - b) (a + b) a 2 -b 2 a*-ab + b* ~ (a 2 -ab + b 2 ) (a + b) ~ a 3 + b 3, **' 78 algesra. EXAMPLES. 2. Reduce to a fraction with a 2 — b 2 for its denom- a + b inator. x -\- 1 3. Reduce - to a fraction with x 2 + 5 x — 24 for its x — 3 denominator. . -r, , 3 m + 2 4. Reduce to a fraction with G m? — 19 m + 10 for its denominator. 4 5. Reduce to a fraction with a 3 — b 3 for its denom- a — inator. 6. Reduce 1 + x to a fraction with 1 — x for its denomi- nator. ADDITION AND SUBTRACTION OF FRACTIONS. 154. 1. Let it he required to add - to -. c c In -- and - , the unit is divided into c equal parts, and a and b parts, respectively, are taken, or in all a + b parts ; that a + 6 is . Ihus, c a b a + b - + - =— -I— . c c c 2. Let it he required to subtract - from - . c c The result must he such a quantity as when added to 7 will produce -; that quantity is evidently ■ • (Art. 154, 1). ,„, a b a — b 1 hus, = . c c c Hence the following FRACTIONS. 79 RULE. To add or subtract fractions, reduce them, if necessary, to a, common denominator. Add or subtract the numerators, and write the result over the common denominator. The final result should be reduced to its simplest form, wherever such reduction is possible. 3b - a b + a . 1 - 4 o 2 1. Add — - , n 7 , and — - — — . 3a ' 2b iab The least common multiple of the denominators is 12 a b. Then, by the rule of Art. 152, 3b -a b + a 1 - 4 b* 12 b 2 - 4 ab 6 a b + 6 a 2 + "ITT" + , , = " -To-T- - + + 3a 2 b ±ab 12 a b 12 a b 3 - 12 lr 12 fr 2 -4 a 6 + 6 a fr + 6 a 2 + 3-12 b 2 12 a b 12 a b 6a 2 +2ab + S 12 a b , Ans. n n i 4 flj — 1 . 6 « — 2 «. subtract — ^ from — . 2 x 6 a The least common denominator is 6 ax. 6a —2 4 a; — 1 12 a a; — 4 a: 12 ax — S a Then, 3a 2a? Gaa; 6 a x 12 a .r — 4 x — (12 ax — 3 a) 12 a a; — 4 a; — 12 a x + 3 a 6 a x 6 ax 3 a — 4 a; 6 a a; , Ans. Note. When a fraction whose numerator is not a monomial is preceded by a - sign, it will be found convenient to enclose its numerator in a pa- renthesis before combining with the other numerators. If this is not done, care must be taken to change the sign of each term in the numerator before combining. 80 ALGEBRA. 4« 2 -l 3«i 2 -2 5« 2 c 2 +3 3. Simplify 2 a c 3 b 2 c 5 a 3 The least common denominator is 30 a b 2 c 3 . 4a 2 -l _ 3 a b 2 - 2 _ 5a 2 c 2 +3 2 a c 3 b 2 c 5 a c 3 60 a 2 b 2 c 2 - 15 b' 2 c 2 _ 30 a 2 b 2 c 2 - 20 a c 2 30 a 2 b 2 c 2 + 18 If 30 a b 2 c s 30 d b 2 c 3 30 a 6 a c 3 _ 60 a 2 b 2 c 2 - 15 b 2 c 2 - (30 a 2 b 2 c 2 -20a c 2 ) - (30 a 2 b 2 c 2 +18 b ? ) 30 a 6 2 c 3 _ 60 a 2 & 2 c 2 - 15 6 2 c 2 - 30 a 2 b 2 c 2 + 20 a c 2 - 30 a 2 b 2 c 2 - 18 b 2 30 a 6' 2 c 3 20 a c 2 - 15 6 2 c 2 - 18 b 2 30 a b 2 c 3 > AnS ' EXAMPLES. Simplify the following : . 2x — 5 3 a + 11 _ a — b 2a + b b — 3a 4. \- . 9. 1 1 . 12 18 4^6 8 3 1 a 2 + 1 6 a 3 + 1 6-2 * 5 a J 2 + 2a 2 i' '3 a 2 12 a 3 + ITT' 2 a + 3 3 a + 5 2a;-l 2x + 3 6cc + l 6 8 ' "12 ~H~ ^2(P" ra — 2 2 — 3?»,?i 2 m + 2 m + 2 m + 3 ' 2m » 3 m 2 n 3 ' ' ~J~ ~U~ ~2lT' b — 4a a + 5b 10 2 2x — 1 3.r 2 +l o. — — 1 — — - — . lo. 24: a ^ 30b ' '3 6x 9 a; 2 ' ,. a — 2 3cc + l 6 a: -5 3 14. ■ -\ ' 2 + 3 4 5 3» + l 2&-1 4<?-l 6^+1 12 a ~~8lT" + 16 c 24^' 16. Simplify FRACTIONS. 81 2x + l 3x — 1 11 2 x (x — 1) 3 a; (a? + 1) 4 (a- 2 - 1) The least common denominator is 12 x (x 2 — 1). 2x + l 3x — 1 11 Then, 2 x (x - 1) 3 x (x + 1) 4 (a; 2 - 1) 6 (a + 1) (2 x + 1) 4 (as - 1) (3 x — 1) 33 a; 12x(x 2 -l) 12x(x 2 -l) 12x(x 2 -l) 12 x 2 + 18 x + 6 12 x- - 16 x + 4 33 x 12 x (x 2 - 1) 12 x (x- - 1) 12 x (x 2 - 1) _ 12 x 2 + 18 x + 6 - (12 x 3 - 16 x + 4) - 33 x 12 x (x 2 - 1) x + 2 12 x (x 2 - 1) Aiis. Simplify the following : »;_* + *_. »i±» + s=» x + 2 3 — x a — 6 a + 6 18. -i L_. 20.^-^. x + 7 x + 8 1 — xl + x a /> 2 a 5 «1. — —7 H j H — o To • a + a — a' — b" 1 1 2x 22. + x + y x — // x' + y- 1 x 3 23. 5 -^ r + 24. X — 1 x 2 — 1 X 3 — 1 2 x — 6 x + 2 x + 1 x 2 +3x + 2 x 2 -2x — 3 x 1 — x— 6' x x 2 x 25. Simplify — — r + — h -5 x + 1 1 — x x' 2 — 1 82 ALGEBRA. The expression may be written (Art. 143) as follows : X X Li X + X + 1 x — 1 X* — 1 The least common denominator is ar — 1. ihen > ZTT^T-Z. T + ~2 7 = 31 i— Z72 T + Z2- x + 1 x — 1 ar — 1 a;' 2 — 1 x' —1 x 2 — 1 or — a: — (.t 2 + cc) + 2 a; x 1 — 1 Simplify the following : x* = 0, Ans. (Art. 102). 3 4 26. -2-+*. 28. -^-_+ — a — & b — a 3 x — a 2 # 2 — 9 _,_. o rt -p x o a — J. -.-. x x x 27. 1 — . 29. 1 3a+3^2-2a 1+x 1-x 1 1 1 30. T-^f-Ti z + x 31. (a — b){b — c) (b — a) (a — c) (c - a) (c — b) 2 3 1 (a; - 2) (a; - 3) (3 -*) (4 -a;) (a; - 4) (2 -a;) ' MULTIPLICATION OF FRACTIONS. 155. We showed, in Art. 137, that a fraction could be multiplied by an integer either by multiplying its numerator or by dividing its denominator by that integer. We will now show how to multiply one fraction by another. Let it be required to multiply - by - . Let - = x, and - = y ; where x and y may be either integral or fractional. FRACTIONS. 83 Since the dividend equals the product of the divisor and quotient. a = b x, and c = d y. Therefore, hy Art. 44, Ax. 3, a c = b d x y. Regarding a c as the dividend, b d as the divisor, and x y as the quotient, we have a c xy = Vd- Therefore, putting for x and y their values, a c a c ~b X d^bd' Hence the following RULE. Multiply the numerators together for the numerator of the resulting fraction, and the denominators for its denominator. Mixed quantities should be reduced to a fractional form before applying the rule. When there are common factors in the numerators and denominators, they should be cancelled before performing the multiplication. EXAMPLES. - ,, 1A . -■ , 6x 2 y 10 a 2 y . 3b*x* 1. Multiply together 5 — ^ , -=-= — - , and -. ~ . 1 J & 5 a 3 b 2 ' 9 b x ' 4 a y 2 6x 2 y 10 a 2 y 3 J 4 x s 6 x 10 X 3 a 2 b* x 5 y 2 b x* . 5a 3 b 2 9bx A lay 2 ' 5 X 9 X 4 « 4 b 3 x y 2 ' ' a 2 ' " Multiply together the following : a 2 5 c -. a 3 b 2 . 3 abx 2 5 x y 2 • — 2 and —3 — j • 4 - -5 — 2- ancl 5 — r - ra ?r m d n d bay 2, ab x 3 a 3 x ,4fflJ _ m ?/ n , a x 3 ' ^li- and KTZ, ■ 5 - 7^1 and 7 A 4 5Am 4 « x m y n 84 ALGEBEA. e 2a 6c ,5b 3ab 2 3ac 2 ,Sa<P o6 5a be 4cd 2&d 9&c „ 8 a-- 15 y 2 ,3 s; 4 _ 3 m* 2 n* . 11 z 2 '• fT~3> i7 — 5 ? and in q • "• ,, o > o — .and-; — 5-. 9 if lb z s ' 10 cc 3 2a; 2 '3??i' 4w 2 10. Multiply together ar 2 _ 2cc x 2 -9 . a 2 + a; , and a;2_2a;-3' a; 2 -*' cc 2 +a:-.6' r-2x x 2 -9 cc 2 + ^ X-s X ce 2 — 2 cc — 3 x- — x x 2 + x — 6 x(x-2)(x + 3) (a - 3) x (x + 1) x (x-3)(x + l)x(x- 1) (a + 3) (x -2) ~ x~^l ' Multiply together the following : 1t 3.r 2 -cc , 10 11. = and 5 2cc 2 -4a;' ,. 4 a; + 2 5cc 1*. — ^ and 2 x + 1 ' 1Q a 2 -2ab + b 2 . b lo. ; and a + b ax — bx ., . a — b . a 2 — b 2 14. -=— -— and a 2 + a b a 2 — a b , _ 1 — x 2 1 — y 2 , 10. q , s , and 1 + 1/ ' X + X 2 ' 1 — X . x 2 -W . x 2 -25 lb. „ — and -5 - — . X s + ox r-4a; a 3 — a 2 + a x s — 8 17. -r and — 5 • x 2 + 2 x + 4 a s + 1 ,_ a: 2 + 5.r+G , x 2 — Ix 18> x^I^^Yl and ^4' FRACTIONS. 85 4 5 x 19. 1 H — and — x x 1 x 2 — 8x + 7 20. -4-1 and «X/ fcC 2 -5cc + 6" a,a _ 3 x + 2 a 2 - 7 a; + 12 a 3 - 5 x 2 2L aj»_8aj + 15' * 2 -5x + 4' and a* -4 ' 22. ^ 2— j Lj ^r> and 1 + ^— a; — x y + y x~ + x y + y x — y a* _ U 1 - c 2 + 2 5 c a 2-p-c 2 -2bc i6 - a - + c--b 2 + 2ac a 2 + c 2 -b 2 -2ac' OA a + b a — b 4 b 2 a + b a — b a + b a — b Jo _„ 2x-\-y . y x 2 .. x 2 — y 2 25. — — — 1 ^ = r- and -= — V- x + y y — cc x* — y* x' + y DIVISION OF FRACTIONS. 156. We showed, in Art. 138, that a fraction could be divided by an integer either by dividing its numerator or by multiplying its denominator by that integer. We will now show how to divide one fraction by another. Let it be required to divide - by - . ft n Let x denote the quotient of --$-—. b a Then, since the quotient multiplied by the divisor gives the dividend, we have c a xe a XX d = b ] ° r ' ~d = b' 86 ALGEBRA. Multiplying each of these equals by - (Art. 44, Ax. 3), G Therefore, a d x = be a c ad d be' Hence the following RULE. Invert the divisor, and proceed as in multiplication. Mixed quantities should he reduced to a fractional form, before applying the rule. EXAMPLES. 6 aH , 9ab 3 1. Divide r . , by - T7 r — = — r. 5 x s y* J 10 x 2 y 5 6a 2 b 9ab 5 6a 2 b 10 x 2 y h Any . : -^ y l_ — "L Ant 5a: 3 ?/ 4 * 10.x 2 ?/ 5 5x s y** 9ab s 3b 2 x' x 2 — 9 x + 3 2. Divide —zrs — by — p — . x 2 — 9 x + S _ (x + 3)(x- 3) 5 _ x — S ~TE~ '' ~5~~~~ ~16~ ~ X x~+3~~ 3~ ' Divide the following: 7 ??i 2 3n 2 x 2 — y 2 x 2 +xy 6 - ~2~ y "13" ' ^-2^ + y 2 y ar-y * . 7« 3 i . 14 a b* _ n Sy 2 5 V 1 J wr ?r o ra w x z — y x — y 5. 18 7??- a: 8 1 6 m 2 a: 4 25 wy 2 } 5« 2 2/ 6 ' -1.1 *■ a 2 +2a-15 } a 2 -2a-3' 6. 1 4t x 2 x 4"^ by 12 + 3- in a; 3 — 4 a* , x 2 —3x-\-2 ' a; 2 + 5x + 6 b} « a +2aj-3' FRACTIONS. 87 COMPLEX FRACTIONS. 157. A Complex Fraction is one having a fraction in its numerator, or denominator, or both. It may be regarded as a case in division, since its numerator answers to the dividend, and its denominator to tbe divisor. However, since multiplying a fraction by any multiple of its denominator must cancel that denominator, to simplify a complex fraction, we may multiphj both of its terms by the least common multiple of their denominators. EXAMPLES. a 1. Reduce - to its simplest form. FIRST METHOD. Proceeding as in division, a c a b a b -=^Xj=—t, Ans. a c a ca b SECOND METHOD. Multiplying both terms by the least common multiple of their denominators, a a -Xbc e c a o . -=- = —T) Ans. da, c a » b xbe a a 2. Reduce — ^— — to its simplest form. b a a — b a + b 88 ALGEBEA. The least common multiple of the denominators is a 2 — b 2 . Multiplying each term by a' 2 — lr, we have a (a -\-b) — a (a — b) cr + ab — a 2 + ab 2 a b Ans. b (a + b) + a (a — b) a b + b~ + a 2 — a b a 1 + b 2 ' 3. Reduce - — to its simplest form. x 1 1 X+ 1 33 + 1 , Ans. 1 1 _, a* a: + 1 + a; 2cc + 1 .1 a; + 1 1 +- Reduce the following to their simplest forms : 4. -L-. 8. !-|. 12 5. b a-\ — c m x n 6. n »- g cc y-^ + 9 7. -5=-4. 11. 15 1 * 1 + X 1 x 2 + X 1 ' 1 + X a b b a 1 1* b a 1 x 2 + 2 X- — 2/ w + ?i 1 «- + ft b + .b 2 ' n 12 x — 7 -\ 9. j. 13. -, a_q 18 a; + 3 X 10. 14. 1 1 — X 1 1 + £C 1 =- -H 1 -= 31 1+ - c + 1 3 — x SIMPLE EQUATIONS. 89 a 2 + b 2 a 2 — lr in — n m 3 — n s , n a 2 — b 2 a 2 + b 2 10 m + n in 3 + n 3 16. ; =— • !"• ~? '?• a + b a — b m + n rnr + nr + a — b a + b m — n mr — n .2 i7. x + y y , i9. •> «2 x + 2 y x 2 " a; v/ a; + y a + x 158. In Art. 42, we defined the reciprocal of a quantity as being 1 divided *by that quantity. Therefore the reciprocal in it of — = — = — ; or, the reciprocal of a fraction is the frac- n in m ii tion inverted. XII. — SIMPLE EQUATIONS. 159. An Equation is an expression of equality between two quantities. Thus, x + 4 = 16 is an equation, expressing the equality of the quantities x + 4 and 16. 160. The First Member of an equation is the quantity on the left of the sign of equality. The Second Member is the quantity on the right of that sign. Thus, in the equation x + 4 = 16, x + 4 is the first member, and 16 is the second member. The sides of an equation are its two members. 161. An Identical Equation is one in which the two mem- bers are equal, whatever values are given to the letters in- volved, if the same value be given to' the same letter in every part of the equation ; as, 90 ALGEBRA. 2a + 2bc = 2(d + bc). 162. Equations usually consist of known and unknown quantities. Unknown quantities are generally represented by the last letters of the alphabet, x, y, z; but any letter may stand for an unknown quantity. Known quantities are repre- sented by numbers, or by any except the last letters of the alphabet. 163. A Numerical Equation is one in which all the known quantities are represented by numbers ; as, 2 x — 11 = x — 5. A Literal Equation is one in which some or all the known quantities are expressed by letters j as, 2x + a = bx 2 — 10. 164. The Degree of an equation containing but one un- known quantity is denoted by the highest power of that unknown quantitj' in the equation. Thus, > are equations of the first degree. and c x = a' 2 + b d ' ) 3 x 2 — 2 x = 65 is an equation of the second degree. In like manner we have equations of the third degree, fourth degree, and so on. When an equation contains more than one unknown quan- tity, its degree is determined by the greatest sum of the exponents of the unknown quantities in any term. Thus, x + x y = 25 is an equation of the second degree. a; 2 — y 2 z = a b 3 is an equation of the third degree. Note. These definitions of degree require that the equation shall not contain unknown quantities in the denominators of fractions, or under radical signs, or affected with fractional or negative exponents. SIMPLE EQUATIONS. 91 165. A Simple Equation is an equation of the first degree. 166. The Root of an equation containing' but one unknown quantity is the value of that unknown quantity; or it is tin- value which, being put in place of the unknown quantity, makes the equation identical. Thus, in the equation 3x — 7 = x + 9, if 8 is put in place of x, the equation becomes 24 - 7 = 8'+ 9, which is identical; hence the root of the equation is 8. Note. An equation may have more than one root. For example, in the equation x 2 = 7 J! -12, if 3 is put in place of x, the equation becomes 9 = 21-12; and if 4 is put in place of x, it becomes 16 = 28 — 12. Each of these results being iden- tical, it follows that either 3 or 4 is a root of the equation. 167 It will be shown hereafter that a simple equation has but one root; an equation of the second degree, two mots; and, in general, that the degree of the equation and the num- ber of its roots correspond. 168. The solution of an equation is the process of finding its roots. A root is verified, or the equation satisfied, when, the root being substituted for its symbol, the equation becomes identical. TRANSFORMATION OF EQUATIONS. 169. To Transform an equation is to change its form with- out destroying the equality. 170. The operations required in the transformation are based upon the general principle deduced directly from the axioms (Art. 44) : 92 ALGEBRA. If the same operations are performed upon equal quantities, the results will be equal. Hence, Both members of an equation may be increased, diminished, multiplied, or divided by the same quantity, without destroy- ing the equality. TRANSPOSITION. 171. To Transpose a term of an equation is to change it from one member to the other without destroying the equality. 172. Consider the equation x — a = &. Adding a to each member (Art. 170), we have x — a + a = b + a or, x = b + a, where — a has been transposed to the second member by changing its sign. 173. Again, consider the equation x + a = b. Subtracting a from each member (Art. 170), we have x + a — a = b — a or, x — b — a, where a has been transposed to the second member by chang- ing its sign. 174. Hence the following EULE. Any term may he indisposed from one member of an equa- tion t<> the other, provided its sign be changed. Also, if the same term appear in both members of an equa- tion affected with tin same sign, it may be suppressed. SIMPLE EQUATIONS. 93 1. In the equation 2x — 12 4- '3 = a; — 5a + 9, transpose the unknown terms to the first member, and the known terms to the second. Eesult, 2x — x + 5 a: = 12 — 3 + 9. EXAMPLES. Transpose the unknown terms to the first member, and the known terms to the second, in the following : 2. 3x — 2a = 45+.2x. 3. 4:X + 9 = 25-12x. 4. 4 a 2 x + b 2 = — 4 a bx+±ac+ b 2 . 5. a c + c x — a d = 2 a — 7 #. 6. & c + a 2 x — m ?r = b x + a tZ — 5. 7. 3 — & — x = c — 3x. 8. 2a — 3c = 5ic — b — dx. 9. 10 jc - 312 = 32 x + 21 - 52 x. CLEARING OF FRACTIONS. 175. 1. Clear the equation -jr T = -X-+ s of fractions. o 4 o o The least common multiple of 3, 4, G, and 8 is 24. Multi- plying each term of the equation by 24 (Art. 170), we have 16 x - 30 = 20 x + 9, where the denominators have been removed. Hence the fol- lowing RULE. Multiply each term of the equation by the least common multiple of the denominators. 2, ax (I x m c — b en x 2 a 1 x :;. 2 a 3 b lab 6 ax ex a 4. x i + , — o b a e 94 ALGEBRA. Note. The operation of clearing of fractions may be performed by- multiplying each term of the equation by any common multiple of the de- nominators. The product of all the denominators is obviously a common multiple, and the rule is sometimes given as follows : "Multiply each term of the equation by the product of all the denominators." EXAMPLES. Clear the following equations of fractions: 3x 5x b 4 3 7. x -%+ 20 = 1 + ^ + 26. n w X O ox n 8 ' is-^-ir +! " l = () - —, *As its %K/ C\C\ f\ *-^ *^ *^ HO J-O </• b ' 5 + L2 = l0 _ "- " 12" 3T~ lo = 8~~6~- 5 x 5 -\^ q x 9J i_j x 10. Clear the equation 21 — ■ — == — — — - - t o 1G 2 of fractions. The least common denominator is 16 ; multiplying each term by 1G, we have 336 - (10 x - 10) = 11 - 3 x - (776 - 56 x) or, : !: U ; - lO x + 10 = 11 - 3 x - 776 + 56 x, Am. Note. When a fraction, whose numerator is not a monomial, is preceded by a —sign, it will he found convenient, on clearing of fractions, to enclose the numerator in a parenthesis. If this is not done, care must be taken to change the sign of each term in the numerator. Clear the following equations of fractions : 11 x a + x 15 12 ax + b cx + d a 2 3~~ ~~2' r be ' V SIMPLE EQUATIONS. 95 13. -1 -1^ = 0. 15. ?_ ?___5a!_ 1 + a; 1 — a? x + 1 a; — 1 ar — 1 14 a ar — 3 1 _a 16 ,T ~*~ ^ a? — 3 2« + l_ 2~2x + l _ 3 _ ' * ~5~ ~2~ ~3~ CHANGING SIGNS. 176. The signs of all the terms of an equation may be changed without destroying the equality. For, in the equation a — x = b — c, let all the terms he multiplied by — 1 (Art. 170). Then, — a + x = — b + c or, x — a = c — b. For example, the equation — 5 x — a = 3 x — b, by chang- ing the signs of all the terms, may he written 5x + a = b — 3x. SOLUTION OF SIMPLE EQUATIONS. 177. To solve a simple equation containing hut one un- known quantity. 1. Solve the equation 5 x — 7 = a? + 9. Transposing the unknown terms to the first member, and the known terms to the second, 5 x — ^ = 7 + 9 Uniting similar terms, 4 x = 16 Dividing each member by 4 (Art. 170), x = 4, Ans. This value of x we may verify (Art. 168). Thus, substi- tuting 4 for x in the given equation, it becomes 20 - 7 = 4 + 9, which is identical ; hence the value of x is verified. 96 ALGEBRA. 2. Solve the equation 8 x + 19 = 25 x — 32. Transposing, 8 x — 25 x = — 19 — 32 Uniting terms, — 17 x = — 51 Dividing by — 17, cc = 3, Ans. To verify the result, put 3 for x in the given equation. Then, 24 + 19 = 75 - 32 or, 43 = 43. o a i .i .- 3x 5 2x x o. bolve the equation — — |- - = — . 4 o *j Clearing of fractions, by multiplying each term of the equa- tion by 12, the least common multiple of the denominators, 9x + 10 = 8x — Gx Transposing, 9a; — 8x + 6x = — 10 Uniting terms, 7 x = — 10 Dividing by 7, x = — — , Ans. To verify this result, put x = =r in the given equation. Then, _30 5_ _20 10 _ .28 + 6~ - 21 + 14 or, or, -90 + 70 _-80 + 60 ~84 _ "SlT _20_ _20 84 " 84' RULE. Clear the equation of fractions if it has any. Transpose the unknown terms to tit e first member, and the known terms to the second, and reduce each member to its simples} firm. Diride both members of the resulting equation by the coefficient of the unknown quantity. SIMPLE EQUATIONS. 97 EXAMPLES. Solve the following equations : 4. 3sc + 5 = a; + ll. 7. 3x + 2-5ce = :b-7 + 3. 5. 3z-2:=5x-lG. 8. 18-5cc-2x = 3 + a; + 7x. 6. 2-2a; = 3-a?. 9. 5x^3 + 17 = 19-2x-2. 10. Solve the equation 5(7 + 3x)-(2a;-3)(l-2x)-(2a ; -3) 2 -(5 + : *0=O- Performing the operations indicated, we have 35 + 15a: + 4a; 2 -8:r + 3-4x 2 + 12:r-9-5-a; = Transposing, and suppressing the terms 4 x 2 and — 4 x 2 , 15 ;c-8;c + 12 a:-a; = - 35 -3 + 9 + 5 18 x = - 24 24 4 , x = -lS = -3> AnS - Solve the following equations : 11. 3 + 2 (2x + S) = 2x -3(2 x + 1). 12. 2cc — (4a;-l)=5a;-0-l). 13. 7 («-2) -5 (a + 3) = 3 (2x- 5) -6 (4a;- 1). 14. 3(3x + 5)-2(5z-3)=13-(5;c-16). 15. (2 x - 1) (3 a; + 2) = (3 a; - 5) (2 a + 20). 16. (5 - G a-) (2 x - 1) = (3 x + 3) (13 - 4 x). 17. (^-3)' 2 -(5-a-) 2 = -4'a;. 18. (2a;-l) 2 -3(^-2) + 5(3x-2)-(5-2a ; j 2 = 0. • 3 7 7 5 19. Solve the equation „ — = zr^ — k~ -• 98 ALGEBRA. Clearing of fractions, by multiplying each term by 12 x, the least common multiple of the denominators, 36 - 42 = 7 x - 20 - 7 x = - 36 + 42 - 20 - 7 a; = - 14 Solve the following equations : 20. i^_7=— -— 24 ^— x -2x 8x 11 4 3 4 ' * 5 *-^--2— -11. 2i 1, 1 1 1 ok x -^ x _x 3x 6 + 2^ _ 4 + 12^- Mt 2 + ~6~3 = 6~7T- «. |-| + | = 18. 26.*-f + 20 = | + | + 2 6. 23. |_?_*=I_i 27. 2-^ = 7- 3 345a? a; 2a; 2a;' oqqt n . • 3a; — 1 2 a; + 1 4a; — 5 ao. holve the equation : = 4. 4 3 5 Multiplying each term by 60, 45 x — 15 - (40 x + 20) - (48 x - 60) = 240 45 x — 15 - 40 x — 20 — 48 x + 60 = 240 45 x - 40 x — 48 x = 15 + 20 - 60 + 240 - 43 a; = 215 X = — 5, ^1?2S. Solve the following equations : oq q 5 ./■ + 3 7x OA 2 .r + 1 r 5 29. 3>x-\ — = -s-. 30. x =— = 5x — -=. 7-2 5 3 31. 7 a- 7 = 3 x + 7. SIMPLE EQUATIONS. 99 32. 2- 7 ^~^ = 3x 33. 34. 6 4 5a--2_3a- + 4 7x-\-2 _x — 10 "IT ~T~ ~6~ ~2~ a; + 1 2 a; — 5 _ 11 x + 5 a; — 13 ~2~ "5" ~W ~3 ' 5a- + l 17 x + 7 3.T-1 7 a; — 1 4 + a- 3 a; — 2 11 a- + 2 2 - 9 a; 36. 14 2 a- + 1 4 a- + 5 8 + x 2x + 5 ' ~3~~ ~T~ ~~6 _ ~8~ 2 3 1 38. Solve the equation x — 1 a 1 + 1 a'~ — 1 " Clearing of fractions, by multiplying each term by x 2 — 1, 2 (a- + 1) - 3 (x - 1) = 1 2a; + 2-3a- + 3 = l 2a-3a; = -2-3 + l — a: = — 4 x = 4, ^4«s. 4 a- + 3 12 a- - 5 2 x - 1 39. Solve the equation 10 5 a; — 1 " 5 Clearing of fractions partially, by multiplying each term by 10, 120 a; -50 , 4a; + 3 =- — t— = ±x — 2 o .'' — 1 _ , . 120 a; -50 4a; + 3 — 4.r + 2 = — = -.— ox — 1 : 120 x - 50 5 = —= — o a- — 1 100 ALGEBRA. Clearing of fractions, by multiplying each term by 5 x — 1, 25 x - 5 = 120 x - 50 25 a; - 120 x = 5 - 50 - 95 £ = - 45 _45_ 9 x -y5-vJ> Ans ' Note. If the denominators are partly monomial, and partly polynomial, clear of fractions at first partially, multiplying by such a quantity as will remove the monomial denominators. Solve the following equations : 1 —X 1 + x 1 — ar x — 1 x + 1 3 X X 2 — 5x 2 3 j3 x - -7 3' 2x-l 2* + 7 3a; + 4~ "3^ + 2 5-2a 3-2x 41 — ' 4fj * ' x ._ 2 s + 2~x 2 -4' ' ~x~+Y'' : ^+T* 9 = "3" " 1-dx ' „_ <6x 2 -3x + 2 _ ._ 2ar + 3a; 1 43 - rr^ s s = 3. 47. -^ — — + — - = x + 1. 2 a; 2 + 5 a; — 7 2 * + 1 3 <c 48. Solve the equation 2 a ce — 3b = x + c — 3 ax. Transposing and uniting terms, 5 a x — x = 3b -\- c Factoring the first member, x (5 a — 1) = 3 b + c Dividing by 5 a — 1, a; = = r , ^4»s. 49. Solve the equation (& — c .r) 2 — (a — c x)' 2 = b (b — a). Performing the operations indicated, b 2 -2bcx+ c 2 x 2 - a 2 + 2acx- c 2 x 2 = b 2 - a b Suppressing the term lr in both members, and the terms c 2 x 2 and — c 2 x 2 in the first member, SIMPLE EQUATIONS. 101 — 2bcx — a 2 + 2acx = — ab 2 a ex — 2b cx = a 2 — ab Factoring both members, 2 c x (a — b) = a (a — b) n . ,. a (a — b) a . Dividing by 2e(a-b), x= 2c ^_^ = ^ , Ans. Solve, the following equations : 50. 2 ax + d — 3c — bx. 51. 2 x — Aa = 3 ax + a 2 — a 2 x. 52. 2 a x + 6 b 2 = 3 b x + 4 a b. 53. 6 b m x — 5 a n — lo a ni — 2bnx. 54. (or - 2 x) 2 = (4 x - b) (x + 4 e). 55. (2 a - 3 x) (2 a + 3x) = b 2 -(3x- b) 2 . 56. (3 a — x) (a + 2 x) = (5 a + x) (a — 2 x), 3b x 2 _ 3 _ 2bx c a c - 3 +4^=3V 2 «( 2 - 3 «>- 57. a X 58. 2a 59. X 2 60. X a b 61. X 2 1 + 2 ax 2x + \ 2 a a 2 X + " b X ~~3lT = 3T~ («-!)• I> c x x a c — 4 6 a; 2b c 6 c 3b c 62. Solve the equation .2 x - .01 - .03 x = .113 x + .161. FIRST METHOD. ('banging the decimals into common fractions. 2x 1 3x 118 a; 161 10 100 100 " 1000 1000 102 ALGEBRA. Multiplying each term by 1000, 200 x - 10 - 30 x = 113 x + 161 57 x = 171 x = 3, Ans. SECOND METHOD. Transposing, .2 x — .03 x — .113 x = .01 + .101 Uniting terms, .057 x = .171 Dividing by .057, x = 3, ^4ms. Solve the following equations : 63. .3x- .02 - .003 x = .7- .06 a- - .006. 64. .001 x - .32 = .09 x - .2 x - .653. 65. .3 (1.2 X -5)=U + .05 x. 66. .7 (x + .13) = .03 (4x- .1) + .5. 67. 3.3a;- \ =.la; + 9.9. .5 2-3a 5a: _ 2x- 3 _ a- - 2 7 "T5 — + L25~ ~9~ "378 ^ " 9 ' 178. To prove that a simple equation ran have but one root. We have soon that every simple equation can be reduced to the form x = a. Suppose, if possible, that a simple equation can have two roots, and that ->\ and r., are the roots of the equation x = a. Then (Art. 168), r x — a, r 2 = a. Hence, t x = r 2 \ that is, the two supposed roots are identical. Therefore a simple equation can have but one root. PROBLEMS. 103 XIII. — PROBLEMS LEADING TO SIMPLE EQUATIONS CONTAINING ONE UNKNOWN QUANTITY. 179. A Problem is a question proposed for solution. 180. The Solution of a problem by Algebra consists of two distinct parts : 1. The Statement, or the process of expressing the condi- tions of the problem in algebraic language, by one or more equations. 2. The Solution of the resulting equation or equations, or the process of determining from them the values of the un- known quantities. The statement of a problem often includes a consideration of ratio and proportion (Art. 21). 181. Ratio is the relation, with respect to magnitude, which one quantity bears to another of the same kind, and is the result arising from the division of one quantity by the other. A Proportion is an equality of ratios. Thus, a : b, or - , indicates the ratio of a to b. a : b = c : d, is a proportion, indicating that the ratio of a to b, is equal to the ratio of c to d. In a proportion the relation of the terms is such that the product of the first and fourth is equal to the product of the second and third. ct c For, a : b = c : <% is the same as j — -, which, by clearing of fractions, gives ad = b c. 1 04 ALGEBRA. 182. For tlie statement of a problem no general rule can be given ; much must depend on the skill and ingenuity of the operator. We will give a few suggestions, however, which will be found useful : 1. Express the unknoivn quantity, <>r one of the unknoivn quantities, by taw of the final letter* of the alphabet. 1'. From tlie given conditions, find expressions for the other unknown quantities, if any, in the problem. 3. Form on equation, by indicating the operations necessary to verify the values of -the unknown quantities, were they already known. 4. Determine the value of the unknown quantity in the equation th US formed. Note. Problems which involve several unknown quantities may often be solved by representing one of them only by a single unknown letter. 1. What number is that to which if four sevenths of itself be added, the sum w.ill equal twice the number, diminished by 27? Let x = the number. 4 x Then -=— = four sevenths of it, and 2x = twice it. 4 x By the conditions, x -\ — — = 2 x — 27 Solving this equation, x = 63, the number required. 2. Divide 144 into two parts whose difference is 30. Let x = one part. Tli en. 144 — x = the other part. By the conditions, x — (144 — x) = 30 Solving this equation, x = 87, one part. 144 — #= ~>7, the other part. PROBLEMS. 105 3. A is three times as old as B ; and eight years ago he was seven times as old as B. What are their ages at present ? Let x = B's age. Then, 3 x — A's age. Now, x — 8 = B's age, eight years ago, and 3 x — 8 = A's age, eight years ago. By the conditions, 3x — 8 = 7 (x — 8) Whence, x = 12, B's age, and, 3 x = 36, A's age. 4. A can do a piece of work in 8 days, which B can perform in 10 days. In how many days can it he done hy both work- ing together ? Let x = the number of days required. Then, - = what both can do in one day. Also, — = what A can do in one day, o and j- = what B can do in one day. Since the sum of what each separately can do in one day is equal to what both can do together in one day, i JL -i 8 + 10 _ x Whence, x = 4f , number of days required. 5. A man has $ 3.64 in dimes, half-dimes, and cents. He has 7 times as many cents as half : dimes, and one fourth as many half-dimes as dimes. How many has he of each ? 106 ALGEBRA. Let x = the number of dimes. x Then, - = the number of half-dimes, 4 7 x and — r— = the number of cents. 4 Now, 10 x = the value of the dimes in cents, and —j— == the value of the half-dimes in cents. 4 By the conditions, 10 x -\ ; — | -— = 364 J '44 Whence, x = 28, number of dimes, x - = 7, number of half-dimes, 4 — t— = 49, number of cents. 6. Two pieces of cloth were purchased at the same price per yard ; but as they were of different lengths, the one cost $ 5 and the other $ 6.50. If each had been 10 yards longer, their lengths would have been as 5 to 6. Required the length of each piece. Since the price of each per yard is the same, the lengths of the two pieces must be in the ratio of their prices, that is, as 5 to 6h, or as 10 to 13. Therefore, Let 10 x = the length of the first piece in yards, and 13 x = the length of the second piece in yards. By the conditions, 10 x + 10 : 13 x + 10 = 5 : 6 or (Art. 181), 6 (10 x + 10) = 5 (13 x + 10) "Whence, x = 2. Then, 10 x = 20, length of first piece, and 13 x = 26, length of second piece. PROBLEMS. 107 7. The second digit of a number exceeds the first by 2 ; and if the number, increased by 6, be divided by the sum of the digits, the quotient is 5. Required the number. Let x = the first digit. Then, x + 2 = the second. Since the number is equal to 10 times the first digit, plus the second, 10 x + x + 2, or 11 x + 2 = the number. 11 x _|_ 2 + 6 By the conditions, ■ ^— = 5 J x + x + 2 Whence, x = 2, the first digit, and x + 2 = 4, the second digit. Therefore the number is 24. 8. Two persons, A and B, 63 miles apart, set out at the same time and travel towards each other. A travels 4 miles an hour, and B 3 miles. What distance will each have trav- elled when they meet ? Let x = the distance A travels. Then, 63 — x = the distance B travels. x - = the time A takes to travel x miles, and — — — = the time B takes to travel 63 — x miles. o By the conditions of the problem, these times are equal ; x 63 — x 4=^r- Whence, x = 36, A's distance, and 63 — x = 27, B's distance. 108 ALGEBRA. 9. At what time between 3 and 4 o'clock are the hands of a watch opposite to each other ? Let M represent the position of the minute-hand at 3 o'clock, and H the position, of the hour-hand at the same time. \jj Let M 1 represent the position of Ih' the minute-hand when it is opposite to the hour-hand, and H 1 the po- sition of the hour-hand at the same time. Let x = the arc M H H' M', the space over which the min- ute-hand has moved since 3 o'clock. x Then, ^ = the arc H H', the space over which the hour- hand has moved since 3 o'clock. Also, the arc MH= 15 minute spaces, and the arc H' M 1 = 30 minute spaces. Now, arc M H H 1 M = arc MH+ arc H H< + arc H< M, x or, x = 15 + j^ + 30 Solving this equation, x = 49 ^ minute spaces. That is, the time is 49-^ minutes after 3 o'clock. PROBLEMS. 10. My horse and chaise are worth $ 336 ; but the horse is worth twice as much as the chaise. Required the value of each. 11. What number is that from which if 7 be subtracted, one sixth of the remainder will be 5? 12. What two numbers are those whose difference is 3, and the difference of whose squares is 51 ? PROBLEMS. 109 13. Divide 20 into two such parts that 3 times one part may be equal to one third of the other. 14. Divide 100 into two parts whose difference is 17. 15. A is twice as old as B, and 10 years ago he was 3 times as old. What are their ages ? 16. A is four times as old as B ; in thirty years he will be only twice as old as B. What are their ages ? 17. A can do a piece of work in 3 days, and B can do the same in 5 days. In how many days can it he done by both working together ? 18. A can do a piece of work in 3§ hours, which B can do in 2| hours, and C in 2i hours. In how many hours can it be done by all working together ? 19. A and B can do a piece of work together in 7 days, which A alone can do in 10 days. In what time could B alone do it ? 20. The first digit of a certain number exceeds the second by 4; and when the number is divided by the sum of the digits, the quotient is 7. What is the number '.' 21. The second digit of a certain number exceeds the first by 3; and if the number, diminished by 9, be divided by the difference of the digits, the quotient is 9. What is the number ? 22. A drover has a lot of oxen and cows, for which he gave $ 1428. For the oxen he gave $ 55 each, and for the cows $ 32 each ; and he had twice as many cows as oxen. Required the number of each. 23. A gentleman, at his decease, left an estate of $1872 for his wife, three sons, and two daughters. His wife was to re- ceive three times as much as either of her daughters, and each son to receive one half as much as each of the daughters. Re- quired the sum that each received. HO ALGEBRA. 24. A laborer agreed to serve for 36 days on these condi- tions, that for every day he worked he was to receive $1.25, but for every day lie was absent he was to forfeit * 0.50. At the end of the time he received $ 17. It is required to find , how many days he labored, and how many days he was absent. 25. A man, being asked the value of his horse and saddle, replied that his horse was worth $114 more than his saddle, and that g the value of the horse was 7 times the value of the saddle. What was the value of each ? 26. In a garrison of 2744 men, there are 2 cavalry soldiers to 25 infantry, and half as many artillery as cavalry. Re- quired the number of each. 27. The stones which pave a square court would just cover a rectangular area, whose length is 6 yards longer, and breadth 4 yards shorter, than the side of the square. Find the area of the court. 28. A person has travelled altogether 3036 miles, of which he has gone 7 miles by water to 4 on foot, and 5 by water to 2 on horseback. How many miles did he travel in each manner ? 29. A certain man added to his estate ^ its value, and then lost $ 760 ; but afterwards, having gained $ 600, his property then amounted to $ 2000. What was the value of his estate at first? 30. A capitalist invested § of a certain sum of money in government bonds paying 5 per cent interest, and the re- mainder in bonds paying 6 per cent ; and found the interest of the whole per annum to be $180. Required the amount of each kind of bonds. 31. A woman sells half an egg more than halt her eggs. Again she sells half an egg more than half her remaining eggs. A third time she does the same; and now she has sold all her eggs. How many had she at first ? PROBLEMS. HI 32. What number is that, the treble of which, increased by 12, shall as much exceed 54, as that treble is less than 144 ? 33. A ashed B how much money he had. He replied, " If I had 5 times the sum I now possess, I could lend you $ 60, and then i of the remainder would be equal to h the dollars I now have." Required the sum B had. 34. A, B, and C found a purse of money, and it was mutu- ally agreed that A should receive $ 15 less than one half, that B should have $13 more than one quarter; and that C should have the remainder, which was $ 27. How many dollars did the purse contain? 35. A number consists of 6 digits, of which the last to the left hand is 1. If tins number is altered by removing the 1 and putting it in the units' place, the new number is three times as great as the original one. Find the number. 36. A prize of $ 1000 is to be divided between A and B, so that their shares may be in the ratio of 7 to 8. Required the share of each. 37. A man has $ 4.04 in dollars, dimes, and cents. He has one fifth as many cents as dimes, and twice as many cents as dollars. How many has he of each ? 38. I bought a picture at a certain price, and paid the same price for a frame ; if the frame had cost $ 1.00 less, and the picture $ 0.75 more, the price of the frame would have been only half that of the picture. Required the cost of the picture. 39. A gentleman gave in charity $ 46 ; a part in equal por- tions to 5 men, and the rest in equal portions to 7 women. Now, a man and a woman had between them $8. What was given to the men, and what to the women ? 40. Separate 41 into two such parts, that one divided by the other may give 1 as a quotient and 5 as a remainder. 112 ALGEBRA. 41. A vessel can be emptied by three taps ; by the first alone it could be emptied in 80 minutes, by the second in 200 minutes, and by the third in 5 hours. In what time will it be emptied if all the taps be opened ? 42. A general arranging his troops in the form of a solid square, finds he has 21 men over; but, attempting to add 1 man to each side of the square, finds he wants 200 men to fill up the square. Required the number of men on a side at first, and the whole number of troops. 43. At what time between 7 and 8 are the hands of a watch opposite to each other ? 44. At what time between 2 and 3 are the hands of a watch opposite to each other? 45. At what time between 5 and 6 are the hands of a watch together ? 46. Divide 43 into two such parts that one of them shall be 3 times as much above 20 as the other wants of 17. 47. Gold is 19} times as heavy as water, and silver 10i times. A mixed mass weighs 4160 ounces, and displaces 250 ounces of water. What proportions of gold and silver does it contain ? 48. A gentleman let a certain sum of money for 3 years at 5 per cent compound interest ; that is, at the end of each year there was added J,, to the sum due. At the end of the third year there was due him .$2315.25. Required the sum let. 49. A merchant ha!s grain worth 9 shillings per bushel 3 and other grain worth 1.'! shillings per bushel, in what proportion must he mix 40 bushels, so that he may sell the mixture at 10 shillings per bushel '.' 50. A alone could perform a piece of work in L2 hours; A and C together could do it in 5 hours; and C's work is § of B's. Now. the work has to be completed by noon. A begins work at 5 o'clock in the morning; at what hour can he he relieved by B and ( '. ami the work- he just finished in time'.' SIMPLE EQUATIONS. H3 51. A merchant possesses $5120, but at the beginning of each year he sets aside a fixed sum for family expenses. His business increases his capital employed therein annually at the rate of 25 per cent. At the end of four years he finds that his capita] is reduced to $3275. What are his annual expenses? 52. At what times between 7 and 8 o'clock are the hands of a watch at right angles to each other ? 53. At what time between 4 and 5 o'clock is the minute- hand of a watch exactly five minutes in advance of the hour- hand ? 54. A person has 11^ hours at his disposal ; how far may he ride in a coach which travels 5 miles an hour, so as to re- turn home in time, walking back at the rate of oh miles an hour ? 55. A fox is pursued by a greyhound, and is 60 of her own leaps before him. The fox makes 9 leaps while the greyhound makes but 6 ; but the latter in 3 leaps goes as far as the former in 7. How many leaps does each make before the greyhound catches the fox ? 56. A clock has an hour-hand, a minute-hand, and a second- hand, all turning on the same centre. At 12 o'clock all the hands are together, and point at 12. How long will it be before the minute-hand will be between the other two hands, and equally distant from each ? XIV. — SIMPLE EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES. 183. If we have a simple equation containing two unknown quantities, as 3 x — 4 y = 2, we cannot determine definitely the values of x and y ; because, for every value which we give to one of the unknown quantities, we can find a corresponding 114 ALGEBRA. value for the other, and thus find any number of pairs of values which will satisfy the given equation. Thus, if we put x = G, then 18 — 4 y = 2, or y = 4 ; t if we put x = — 2, then — 6 — 4 y = 2, or y = — 2 ; if we put a; = 1, then 3 — 4 ?/ = 2, or y = £ ; etc. And any of the pairs of values < " , I, ■! n i, ■< ~ , , etc., will satisfy the given equation. If we have another equation of the same kind, as 5x + 7?/=17, we can find any number of pairs of values which will satisfy this equation also. Now suppose we are required to determine a pair of values which will satisfy both equations. We shall find but one pair of values in this case. For, multiply the first equation by 5 ; thus, 15 x -20 y = 10; and multiply the second equation by 3 ; thus, 15 x + 21 y = 51. Subtracting the first of these equations from the second (Art. 44), we have 41 y = 41, or, p = l. In the first given equation put y = l; then 3 x — 4 = 2, or 3 x = 6 ; whence, x = 2. The pair of values \ ~-i\ satisfies both the given equations; and no other pair of values can be found which will satisfy both. 184. Simultaneous Equations are such as are satisfied by the same values of their unknown quantities. 185. Independent Equations are such as cannot be made to assume the same form. SIMPLE EQUATION'S. 115 186. It is evident, from Art. 183, that two unknown quantities require for their determination two independent, simultaneous equations. When two such equations are given, it is our object to obtain from them a single equation contain- ing but one unknown quantity. The value of that unknown quantity may then be found; and by substituting it in either of the given equations we can find, as in Art. 183, the value of the other. ELIMINATION". 187. Elimination is the process of combining simultaneous equations so as to obtain from them a single equation contain- ing but one unknown quantity. There are four principal methods of elimination : by Addi- tion or Subtraction, by Substitution, by Comparison, and by Undetermined Multipliers. CASE I. 188. Ellin {nation by Addition or Subtraction. 1. Given ox — 3 y = 19, and 7 x + 4 y — 2, to find the values of x and y. Multiplying the first equation by 4, 20 x — 12 y = 76 Multiplying the second equation by 3, 21 x + 12 y = 6 Adding these equations, 41 x = 82 Whence, x = 2. Substituting this value in the first given equation, 10-3y = 19 -3y = 9 y = -3. We might have solved the equations as follows : Multiplying the first by 7, 35 x - 21 y = 133 (1) Multiplying the second by 5, 35 x + 20 y = 10 (2) Subtracting (2) from (1), — 41 y = 123 2/ = -3. 116 ALGEBRA. Substituting this value of y in the first given equation, 5 x + 9 = 19 5 a- = 10 x = 2. The first of these methods is elimination by addition ; the second, elimination by subtraction. RULE. Multiply the given equations, if necessary, by such numbers or quantities as will make the coefficient of one of the unknown quantities the same in the two resulting equations. Then, if the signs of the terms having the same coefficient arc alike, subtract one equation from the other ■ if unlike, add the two equations. This method of elimination is usually the best in practice. CASE II. 189. Elimination by Substitution. Taking the same equations as before, 5 x — 3 y = 19 (1) 7x + 4t/= 2 (2) Transposing the term 7 x in (2), 4 y = 2 — 7 x 2 7 x Dividing by 4, y = — _ (3) Substituting this value of y in (1), 5*_3 (^=^) =19 Performing the operations indicated, SIMPLE EQUATIONS. 117 Clearing of fractions, 20 x — (6 — 21 x) = 76 or, 20 x - 6 + 21 x = 76 Transposing, and uniting terms, 41 x = 82 Whence, x = 2. 2 — 14 Substituting this value in (3), y = — j — = — 3. • RULE. 7v//r/ £/ie 7v//»e o/o/^e o/ £Ae unknown quantities in terms of the other, from cither of the given equations; and substi- tute this value for that quantity in the other equation. This method is advantageous when either of the unknown quantities has 1 for its coefficient. CASE III. 190. Elimination by Comparison. Taking the same equations as before, 5 x - 3 y = 19 (1) 7x + ±y= 2 (2) Transposing the term — 3 y in (1), 5 x = 3 y + 19 3//+ 19 ,,. or, a = — g (3) Transposing the term 4 y in (2), 7 a; = 2 — 4 y or, a: = — Placing these two values of x equal to each other (Art. 44), 3// + 19 _ 2-4y 5 7 Clearing of fractions, 21 y + 133 = 10 — 20 y 118 ALGEBRA. Transposing, and uniting terms, 41 y = — 123 Whence, y — — 3. — 9 + 19 Substituting this value in (3), x = p o RULE. Find the value of the same unknown quantity in terms of the other, from each of the given equations : ami form a new equation by placing these values equal to each other. CASE IV. 191. Elimination by Undetermined Multipliers. An Undetermined Multiplier is a factor, at first undeter- mined, but to which a convenient value is assigned in the course of the operation. Taking the same equations as before, 5x-3y = 19 (1) 7 x + 4 y = 2 (2) Multiplying (1) by m, 5 m x — 3 m y = 19 m (3) Subtracting (3) from (2), 7 x — 5mx + iy + 3m y — 2 — 19 m Factoring, x (7 — 5 m) + y (4 + 3 m) =2 — 19 m (4) Now, let the coefficient of y, 4 + 3 m = ; then 3 m = — 4, 4 or m = — Kj substituting this value of m in (4), o /_ 20\ „ 76 n 7 +ir) = 2 +3 Clearing of fractions, x (21 + 20) = 6 + 70 41 x = 82 x = 2. SIMPLE EQUATIONS. 119 Substituting this value in (2), 14 + 4 y = 2 4y = -12 y = -3. We might liave let the coefficient of x in (4), 7 — 5m = 0; 7 then m would have been ■= ; substituting this value of m in (4), o y(±+ )=2-- Clearing of fractions, y (20 + 21) = 10 - 133 41 y = - 123 y = -3. Instead of subtracting (3) from (2), we migbt have added them and obtained the same results. Also, in the first place, we might have multiplied (2) by m, and either added the re- sult to, or subtracted it from, (1). RULE. Multiply one of the given equations by the undetermined quantity, m ; and add the result to, or subtract it from, the other given equation. In the resulting equation, factored with reference to the unknown quantifies, place the coefficient of one of the un- known quantities equal to zero, and find the value of ra. Substitute tins value of m In the equation, and the result trill be a simple equation containing but one unknown quantity. This method is advantageous in the solution of literal equations. 2. Solve the equations. ax + b y — c (1) a f x + b'y = c' (2) 120 ALGEBRA. Multiplying (1) by m, a m x + b m y = c m (3) Add (2) and (3), a' x + a m x + V y + b m y = d + cm Factoring, x (a' + a vi) + y (b 1 + b m) = d + c m (4) In (4), put the coefficient of y, V + b in, equal to zero. Then, b m — — V ; whence, m = — -. b Substituting this value of m in (4), a V \ , c b x \a' — ) = d Clearing of fractions, x (a' b — ab') =b d — b' e 1, ( j _ y c Whence, x = — — . a' b — ab' In (4), put the coefficient of x, a' + a m, equal to zero. Then, a m = — a ' ; whence, m = ■ . a Substituting this value of m in (4), a'b\ a'e y\ h — '-) — c — a ' a Clearing of fractions, y (ab 1 — a 1 b) = ad — a' c a d — a' c Wh ence, y = —r l — . ab' — a' b Before applying either of the preceding methods of elimina- tion, the given equations should be reduced to their simplest forms. EXAMPLES. 192. Solve, by whichever method may be most advanta- geous, the following equations : 3. 3cc + 72/ = 33; 2x + ±y = 20. 4. 7x + 2y = 31; 3 z ~ 4 ?/ = 23. 5. 6x-3y = 27; 4;r-6y = -2. SIMPLE EQUATIONS. 121 6. 7 x + 3y = -50; 2y-5x = U. 7. 8y + 12 .£ = 116 ; 2x — y = 3. 8. 11 x + 3 y = - 124 ; 2 x - 6 y = 56. 9. 9a; + 4?/ = 22;27/ + 3cc = 14. 10. ^+^f-8; -8x + 2y = -S0. 11. 7z-2y = G;2a; + 22/ = -24. ,«-.- ^ ,.„ 2^; 11?/ 5 12. 11 y + G a; = 115 ; — — = — ■= . 13. | a} + |y = ^ ; 10*-12y = -62. 5 7 14. _7a; + 47/ = -113; £ + -?, = -. 15> 2~3-°' 4 + G- b - 16. ^- y = 31; a; + ^ = 33. 17. A^_^ = _30; aj + 7y = 119. < 3 a 18. rc + 2// = .G; 1.7 x -y = . 58. 3 ^ ?/ cc 2 ?/ 19. -t r -?-? 1[g L=^.4y-8 aJ = lL 2 T » + 3y 3 7 y - x ' 2x-y 8' 2 + x + 2y ~ 21. a x + b y = m ; c x + d y= n. 22. m cc + ?i ?/ = r ; m' cc — n' y = /. no xy y x aw a c 122 ALGEBRA. x v 1 x y 1 ** a + b^ a~b a 2 -b 2> a-b^ a + b a 2 -b 2 9* X 19 Vm*. X S 2 V~ X -Q7 X + V 25l -2~ 12 = 4 +8 ' 3~ 8 4T~~^ V 2 2,3 2 26. + = 1;— - = 0. 03 + // X- — V/ 03 + 2/ ^ — i7 2x + y _17 2y + 03 4 2 a? - y _ 2 y - a? 4*. x- -g- - 12 4 "' 3 4 "-y " 3 ' 2 03 3.7/ x + 2?/_ 5x — 6y 28< "3 5 4~~ ~° 4~"' 03 ox — y H x 29. Solve the equations, 6_3_ x y~ 8 15 - + — = -1 « y Multiplying the first equation hy 5, 30_15 = 2Q x y ' Adding this to the second given equation, " = 19 03 Clearing of fractions, 38 = 19 03 Whence 03 = 2. Substituting this value in the first given equation, y Transposing, = 1 Whence, y = — 3. SIMPLE EQUATIONS. 123 Solve the following equations : 3 15 2 3 -1. 30. x y 4 ' x y 31. 12 18 42 15 8 17 x y 5 ' x y 3 32. 11 7 3 4 8 x y 2 ' x y -10. nn a b c d 66. - + - = m ; -+- = ». x y x y fi 4 9 8 34. — + f- = 4a5; -^ — = 3a 2 -4J a . <za; 6y 6a; ay oc m ?i n m ,o oO. 1 =zm + n; — | = iiv + n". nx my x y XV. — SIMPLE EQUATIONS CONTAINING MORE THAN TWO UNKNOWN QUANTITIES. 193. If we have given three independent, simultaneous equations, containing three unknown quantities, we may coin- hine two of them hy the methods of elimination explained in the last chapter, so as to obtain an equation containing only two unknown quantities ; we may combine the third equation with either of the two former in the same way, so as to obtain another equation containing the same two unknown quantities. Then from these two equations containing two unknown quan- tities we may derive, as in the last chapter, the values of those unknown quantities. These values being substituted in either of the given equations, the value of the third unknown quan- tity may be determined from the resulting equation. The method of elimination by addition or subtraction is usually the most convenient. 124 ALGEBRA. 194. 1. Solve the equations, 8x-dy-7z = -36 12 x— y — 3z — 36 6x-2y- z = 10 Multiplying the first by 3, 21 x — 27 y - 21 z = — 108 (1) Multiplying the second by 2, 21 x — 2 y — 6 z = 72 (2) Multiplying the third by 4, 21 x — Sy — 4tz = 10 (3) Subtracting (1) from (2), or, Subtracting (3) from (2), or, Multiplying (5) by 3, Adding (1) and (6), Substituting this value in (5), Substituting the values of y and z in the third given equation, sc = 4. In the same manner, if we have given n feidependent, simultaneous equations, containing n unknown quantities, we may combine them so as to form a — 1 equations, containing n — 1 unknown quantities. These, again, may be combinnl so as to form n — 2 equations, containing n — 2 unknown quantities; and so on : the operation being continued until we finally obtain one equation containing one unknown quantity. RULE. Multiply the given equations, if necessary, by such numbers or quantities as trill make the coeffirirnt of one of the un- known quantities the same in the resulting equations. Cont- inue these equations by addition <>r subtraction, so as to form 2oy + loz = ISO 5y+ 3z= 36 (4) 6y- 2z= 32 3y- z= 16 (5) 9y- 3z= 18 (6) 11//= 81 y = 6. z = 2. SIMPLE EQUATIONS. 125 a new set of equations, one less in number than before, and containing one less unknown quantity. Continue the opera- tion with these new equations ; and so on, until an equation is obtained containing nut- unknown quantity. Find the value of this unknown quantity. By substituting it in either of the equations containing only two unknown quantities, find the value of a second unknown quantity. By substituting these values in either of the equations containing three unknown quantities, find the value of a third unknown quantity j and so on, until the values of all arc found. Note. This rule corresponds only with the method of elimination by- addition or subtraction ; which, however, as we have observed before, is the best in practice. 2. Solve the equations, u + x + y = 6 u + x -\- z = 9 u + y + z = 8 x+y+z=7 The solution may here be abridged by the artifice of assum- ing the sum of the four unknown .quantities to equal an auxil- iary quantity, s. Thus, Let u + x + y + z = s. Then we may write the four given equations as follows : s-z = 6 (1) s-y=9 (2) s-x = S (3) s-u=7 (4) Adding, 4 s — s = 30 Whence, s = 10. Substituting the value of s in (1),,(~), (3), and (4), we obtain z ■= 4, y = 1, x = 2, and u = 3. 126 ALGEBRA. EXAMPLES. Solve the following equations : 3. x + y+z = 53; x + 2 y + 3s = 107; x + 3y + 4s = 137. 4. 3x-y — 2z = -23\ G x + 2 y + 3 z = 15; 4x + 3 y — z = — G. 5. 5x — 3y+2z = 41i 2x + y-z = l7; 5x + ±y-2z = 3%. 6. 7a; + 47/-2 = -50; 4x- 5 y-3 z = 20; x — 3y — 4:Z = 30. 7. 3u + x + 2y — z = 22; 4x — y + 3 s = 35; 4:u + 3x-2y=19; 2 u + 4y + 2 2 = 46. 8. a; + ?/ = 2j a2 + s = 3; y + * = — 1. 9. ?/ + s = a ; aj + £ = J ; a; + y = c . 10. 4a;-4y = a + 4s; 6y — 2a; = a + 2s; 7s-y = fl + ^. 11 2 + 3~4-~ 43 ' 3"I + 2-° 4 ' 4 + 2~3 == ~°°- 12. 2a + 2 2 , + 3 = -17-2w; ?/ + 3*=-2; 4aj. + * = 13j ^ + 3y = -14. 13. a y -\- b x = c; c x + a z = b ; b z + c y = a. 14 4 ?_ 6 81 _2_ _3_ 10 _7 a: y 2~ = 2~' 3^" + 2l/~7 r i" Z "2 8 6 4 .. + — = 11. 9 .'■ // 7 .- 3 2 J_ J_ -1 A A -1 °' 4x "37/" ; ' ""32/ + 2z~ ; 2z + 4:x~ 16. x — ay+ cr z — a 8 ; x — by+b' 2 z = b :i ; x — c y + c' 2 z = c 8 PROBLEMS. 127 17 y~ z x + % _ 1 x —y x — z —n. 2 ~ 4 ~2' 5 6 y+z x + y 4 2 -4. 18. £^±1_ (2 _,) = 0; ^±-^ = 2a-cx a c (« + f) 2 -ac(2 + a; + s) = -y. XVI. — PROBLEMS LEADING TO SIMPLE EQUATIONS CONTAINING MORE THAN ONE UNKNOWN QUANTITY. 195. In the solution of problems in which we represent more than one of the unknown quantities by letters, we must obtain, from the conditions of the problem, as many indepen- dent equations as there are unknown quantities. 1. If 3 be added to both numerator and denominator of a certain fraction, its value is f ; and if 2 be subtracted from both numerator and denominator, its value is h- Required the fraction. Let and By the conditions, X- = the numerator, y-- = the denominator, X + 3 o V + 3" o O X o w 1 y -2 2 x ■■ = ~> y = 12. Solving these equations, That is, the fraction is T v. 2. The sum of the digits of a number of three figures is 13 ; if the number, decreased by 8, be divided by the sum of the second and third digits, the result, is 25; and if 99 be added to the number, the digits will be inverted. Find the number. 128 ALGEBRA. Let x = the first digit, y = the second, and z — the third. Then, 100 x + 10 y + z = the nuniher, and 100 z + 10 y + x = the number with its digits inverted. By the conditions, x + y + z = 13 100 x + 10 y + z-8 — Zo 100 x + 10 y + z + 99 = 100 z + 10 y + x Solving these equations, x = 2, the first digit, y = 8, the second, 3, the third. That is, the number is 283. 3. A crew can row 20 miles in 2 hours down stream, and 12 miles in 3 hours against the stream. Required the rate per hour of the current, and the rate per hour of the crew in still water. Let x = rate per hour of the crew in still water, and y = rate jjer hour of the current. Then, x + y= rate per hour rowing down stream, and x — y = rate per hour rowing up stream. Since the distance divided by the rate gives the time, we have by the conditions, 20 2 x + y 12 = 3 x—y Solving these equations, x = 7, and y = 3. PROBLEMS. 129 PROBLEMS. 4. A says to B, " If i of my age were added to § of yours, the sum would be 19^ years." " But," says B, " if § of. mine were subtracted from J of youx-s, the remainder would be 18£ years." Required their ages. 5. If 1 be added to the numerator of a certain fraction, its value is 1 ; but if 1 be added to its denominator, its value is £. What is the fraction? 6. A farmer has 89 oxen and cows ; but, having sold 4 oxen and 20 cows, found he then had 7 more oxen than cows. Re- quired the number of eacb at first. 7. A says to B, " If 7 times my property were added to \ of yours, the sum would be % 990." B replied, " If 7 times my property were added to \ of yours, the sum would be $ 510." Required the property of each. 8. If \ of A's age were subtracted from B's age, and 5 years added to the remainder, the sum would be 6 years ; and if 4 years were added to \ of B's age, it would be equal to fe of A's age. Required their ages. 9. Divide 50 into two such parts that % of the larger shall be equal to § of the smaller. 10. A gentleman, at the time of his marriage, found that his wife's age was to his as 3 to 4 ; but, after they had been mar- ried 12 years, her age was to his as 5 to 6. Required their ages at the time of their marriage. 11. A farmer hired a laborer for 10 days, and agreed to pay him $ 12 for every day he labored, and he was to forfeit % 8 for every day he was absent. He received at the end of his time % 40. How many days did he labor, and how many days was he absent ? 12. A gentleman bought a horse and chaise for % 208, and i of the cost of the chaise was equal to § the price of the horse. What was the price of each ? 130 ALGEBRA. 13. A and B engaged in trade, A with $ 240, and B with $96. A lost twice as much as B; and, upon settling their accounts, it appeared that A had three times as much remain- ing as B. How much did each lose ? 14. Two men, A and B, agreed to dig a well in 10 days; hut, having labored together 4 days, 1! agreed to finish the job, which he did in 16 days. How long would it have taken A to dig the whole well ? 15. A merchant has two kinds of grain, one at 60 cents per bushel, and the other at 90 cents per bushel, of which he wishes to make a mixture of 40 bushels that may be worth 80 cents per bushel. How many bushels of each kind must he use '.' 16. A farmer has a box filled with wheat and rye; seven times the bushels of wheat are 3 bushels more than four times the bushels of rye ; and the quantity of wheat is to -the quan- tity of rye as 3 to 5. Required the number of bushels of each. 17. My income and assessed taxes together amount to $ 50. But if the income tax be increased 50 per cent, and the as- sessed tax diminished 25 per cent, the taxes will together amount to $ 52.50. Required the amount of each tax. 18. A and B entered into partnership, and gained 8200. Now 6 times A's accumulated stock (capital and profit) was equal to 5 times B's original stock; and 6 times B's profit exceeded A's original stock by $200. Required the original stock of each. 19. A boy at a fair spent his money for oranges. If he had got five more for his money, they would have averaged a half- cent less ; and if three less, a half-cent each more. J low many cents did he spend, and how many oranges did he get? 20. A merchant has three kinds of sugar. He can sell 3 lbs. of the first quality, 4 lbs. of the second, and 2 lbs. of the third, for 60 cents; or, he can sell 4 lbs. of the first quality, 1 lb. of the second, and 5 lbs. of the third, for 59 cents ; or, he PROBLEMS. 131 can sell 1 lb. of the first quality, 10 lbs. of the second, and 3 lbs. of the third, for 90 cents. Required the price per lb. of each quality. 21. A gentleman's two horses, with their harness, cost him $120. The value of the poorer horse, with the harness, was double that of the better horse; and the value of the better horse, with the harness, was triple that of the poorer horse. What was the value of each ? 22. Find three numbers, so that the first with half the other two, the second with one third the other two, and the third with one fourth the other two, shall each be equal to 34. 23. Find a- number of three places, of which the digits have equal differences in their order ; and, if the number be divided by half the sum of the digits, the quotient will be 41 ; and, if 396 be added to the number, the digits will be inverted. 24. There are four men, A, B, C, and D, the value of whose estates is $14,000; twice A's, three times B's, half of C's, and one fifth of J)'s, is $16,000; As, twice B's, twice C's, and two fifths of D's, is $18,000; and half of A's, with one third of B's, one fourth of C's, and one fifth of D's, is $ 4000. Re- quired the property of each. 25. A and B are driving their turkeys to market. A says to B, " Give me 5 of your turkeys, and I shall have as many as you." B replies, " Give me 15 of yours, and then yours will be f of mine." How man}' had each ? 26. A says to B and C, " Give me half of your money and I shall have $ 55." B replies, " If you two will give me one third of yours, I shall have $ 50." But C says to A and B, " If I had one fifth of your money I should have $50." Required the sum that each possessed. 27. A gentleman left a sum of money to be divided among his four sons, so that the share of the eldest was \ of the sum of the shares of the other three, the share of the second J of the sum of the other three, and the share of the third I of the 132 ALGEBRA. sum of the other three ; and it was found that the share of the eldest exceeded that of the youngest by $ 14. "What was the whole sum, and what was the share of each person? 28. If I were to enlarge my field by making it .") rods longer and 1 rods wider, its area would be increased by 240 square rods; but if I were to make its length 4 rods less, and its width 5 rods less, its area would be diminished by 210 square rods. Required the present length, width, and area. 29. A boatman can row down stream, a distance of 20 miles, and back again in 10 hours; and he finds that he can row 2 miles against the current in the same time that he rows 3 miles with it. Required the time in going and in returning. 30. A and B can perform a piece of work in days. A and C in 8 days, and B and C in 12 days. In how many "lays can each of them alone perform it ? 31. A person possesses a capital of $30,000, on which he gains a certain rate of interest; but he owes $20,000, for which he pays interest at another rate. The interest which he receives is greater than that which he pays by $800. A second person has $35,000, on which he gains the second rate of interest ; but he owes $ 24,000, for which he pays the first rate of interest. The sum which he receives is greater than that which he pays by $ 310. What are the two rates of in- terest ? 32. A man rows down a stream, which runs at the rate of o\ miles per hour, for a certain distance in 1 hour and 40 min- utes. In returning it takes him 6 hours and .'50 minutes to arrive at a point 2 miles short of his starting-place. Find the distance lie pulled down the stream, and the rate of his pulling. 33. A train running from Boston to New York meets with an accident which causes its speed to be reduced to ,1 of what it was before, and it is in consequence 5 hours late. If the accident bad happened 60 miles nearer New York, the train would have been only one hour late. "What was the rate of the train before the accident ? PROBLEMS. 133 34. A and B run a mile. A gives B a start of 44 yards and beats him by 51 seconds, and afterwards gives him a start of 1 minute 15 seconds and is beaten by 88 yards. In how many minutes can each run a mile ? 35. A merchant has two casks, each containing a certain quantity of wine. In order to have an equal quantity in each, he pours out of the first cask into the second as much as the second contained at first ; then he pours from the second into the first as much as was left in the first; and then again from the first into the second as much as was left in the second, when there are found to be 1G gallons in each cask. How many gallons did each cask contain at first ? 36. A and B arc building a fence 12G feet long; after three hours A leaves off, and B finishes the work in 14 hours. If seven hours had occurred before A left off, B would have fin- ished the work in 4§ hours. How many feet does each build in one hour ? GENERALIZATION OF PROBLEMS. 196. A problem is said to be generalized when letters are used to represent its known quantities, as well as unknown. The unknown quantities thus found in terms of the known are general expressions, or formula', which may be used for the solution of any similar problem. 197. The algebraic solution of a generalized problem dis- closes many interesting truths and useful practical rules, as may be seen from the consideration of the following : 1. The sum of two numbers is a, and their difference is b ; what are the two numbers ? Let x = the greater number, and y = the less. By the conditions, x + y==a x — y = b 134 ALGEBRA. , . , . a + b . , ►solving these equations, x = — - — , the greater number, i a — b , , and y = — - — , the less. Hence, since a and b may have any value 'whatever, the values of x and y are general, and may be expressed as rules for the numerical calculations in any like case; thus. To find two numbers when their sum and difference are given, — Add the sum and difference, and divide by 2, for the greater of tin 1 two numbers; and subtract the difference from the sum, and divide by 2, for the less number. For example, if the sum of two numbers is 35, and their difference 13, the greater = — = 24, and the less = - — - — = 11. 2. A can do a piece of work in a days, which it requires b days for B to perform. In how many days can it be done if A and B work together ? Let x = the number of days required. Then - = what both together can do in one day. Also, - = what A can do in one day, and y = what B can do in one day. By the conditions, - + - = - a b x Whence, x — - — , number of days required. Hence, to find the time for two agencies conjointly to ao- PROBLEMS. 135 complish a certain result, when the times are given in which each separately can accomplish the same, — Divide the product of the given tunes by their sum. For example, if A can do a piece of work in 5 days, and B in 4 days, the time it will take them hoth working together •ill 5x4 20 o, , will be -z . = — = Z% days. 5 + 4 9 J J 3. Three men, A, B, and C, enter into partnership for a certain time. Of the capital stock, A furnishes m dollars; B, n dollars; and C, p dollars. They gain a dollars. "What is each man's share of the gain ? Let x = A's share. Then, since the shares arc proportional to the stocks, n ./• in = B's share, and == C's share. m ix x ty %c By the conditions, x -\ 1 — = a in in Whence, x = ■ , A's share. /// + n + p Then, = , B's share, m m + n + p * and — — = , C's share. m in + n + p Hence, to find each man's gain, when each man's stock and the whole gain are given, — Multiply the whole gain by each man's stock, and divide tlisproduct by the whole stock. For example, suppose A's stock $300, B's $500, and C's $800, and the whole gain $320. 136 ALGEBRA. Then, A's share 320 x 300 96000 = $60, "300 + 500 + 800" 1600 B's share 320 x 500 160000 = $100, "300 + 500 + 800" 1600 and C's share 320 x 800 256000 = $160. 300 + 500 + 800 1600 PROBLEMS. 4. A cistern can he rilled by three pipes ; bj the first in a hours, by the second in b hours, and by the third in c hours. In what time can it he filled by all the pipes running together ? 5. Using the result of the previous problem, suppose that the first pipe fills the cistern in 2 hours, the second in 5 hours, and the third in 10 hours. In what time can it be filled by all the pipes running together ? 6. Divide the number a into two parts which shall have to each other the ratio of m to n. 7. Using the result of the previous problem, divide the number 20 into two parts which shall have to each other the ratio of 3 to 2. 8. A courier left this place n days ago, and goes a miles each day. He is pursued by another, starting to-day and going b miles daily. How many days will the second re- quire to overtake the first ? 9. In the last example, if n = 3, a — 40, and b = 50, how many days will he required ? 10. Required what principal, at interest at r per cent, will amount to the sum a, in t years ? 11. Using the result of the previous problem, what principal, at 6 per cent interest, will amount to $3108 in 8 years? J.2. Required the number of years in which j> dollars, at r per cent interest, will amount to a dollars. DISCUSSION OF PROBLEMS. 137 13. Using the result of the previous problem, in how many years will $262, at 7 per cent interest, amount to $472.91 ? 14. A banker has two hinds of money. It takes a pieces of the first to make a dollar, and b pieces of the second to make the same sum. If be is offered a dollar for c pieces, how many of each kind must he give ? 15. In the last example, if a = 10, b = 20, and c= 15, how many of each kind must he give ? 16. A gentleman, distributing some money among beggars, found that in order to give them a cents each he should want b cents more; he therefore gave them c cents each, and had d cents left. Required the number of beggars. 17. A mixture is made of a pounds of coffee at m cents a pound, h pounds at n cents, and c pounds at j> cents. Re- quired the cost per pound of the mixture. 18. A, B, and C hire a pasture together for a dollars. A puts in m horses for t months, B puts in n horses for t' months, and C puts in^j horses for t" months. What part of the ex- pense should each pay ? XVII. — DISCUSSION OF PROBLEMS LEADING TO SIMPLE EQUATIONS. 198. The Discussion of a problem, or of an equation, is the process of attributing any reasonable values and relations to the arbitrary quantities which enter the equation, and inter- preting the results. 199. An Arbitrary Quantity is one to which any reason- able value may be given at pleasure. 200. A Determinate Problem is one in which the given conditions furnish the means of finding the required quantities. 138 ALGEBRA. A determinate problem leads to as many independent equa- tions as tin-re are required quantities (Art. 195). 201. An Indeterminate Problem is one in which there are fewer imposed conditions than there are required quantities, and, consequently, an insufficient number of independent equations to determine definitely the values of the required quantities. 202. An Impossible Problem is one in which the condi- tions are incompatible or contradictory, and consequently can- not be fulfilled. 203. A determinate problem, leading to a simple equation involving only one unknown quantity, can be satisfied by but one value of that unknown quantity (Art. 178). An indeterminate problem, or one leading to a less number of independent equations than it has unknown quantities, may be satisfied by any number of values. For example, suppose a problem involving three unknown quantities leads to only two equations, which, on combining, give x — z = 10, or, x = 10 + z. Now, if we make z = 1, then x = 11 ; z — 2, then x = 12 ; z = 3, then x = 13. Thus, we may find sets of values without limit that will sat- isfy the equation. Hence, An indeterminate equation may have any manlier of so- lutions. 204. When a problem leads to more independent equations than it lias unknown quantities, it is impossible. For, suppose we have a problem furnishing three indepen- dent equations, as, x = y+l y — 7 — x xy = 16 DISCUSSION OF PROBLEMS. 139 From the first two we find x = 4 and y = 3. But the third requires their product to be 10; hence the problem is im- possible. If, however, the third equation had not been independent, but derived from the other two, as, x y = 12, then the problem would have been possible; but the last equa- tion, not being required for the solution, would have been redundant. INTERPRETATION OF NEGATIVE RESULTS. 205. In a Negative Result, or a result preceded by a — sign, the negative sign is regarded as a symbol of interpre- tation. Its significance when thus used it is now proposed to in- vestigate. 1. Let it be required to find what number must be added to the number a that the sum may be b. Let x = the required number. Then, a + x = b Whence, x = b — a. Here, the value of x corresponds with any assigned values of a and b. Thus, for example, Let a = 12, and b = 25. Then x = 25 - 12 = 13, which satisfies the conditions of the problem ; for if 13 be added to 12,* or a, the sum will be 25, or b. But, suppose a = 30, and b = 24. Then, x = 24 - 30 = - 0, 140 ALGEBRA. which indicates that, under the latter hypothesis, the problem is impossible in an arithmetical sense, though it is possible in the algebraic sense of the words "number," "added," and " sum." The negative result, — 6, points out, therefore, in the arith- metical sense, either an error or "// impossibility. But, taking the value of x with a contrary sign, we see that it will satisfy the enunciation of the problem, in an arithmeti- cal sense, Avhen modified so as to read : What number must be taken from 30, that the remainder may be 24 ? 2. Let it be required to determine the epoch at which A"s age is twice as great as B's ; A's age at present being 35 years, and B's 20 years. Let us suppose the required epoch to be after the present date. Let x — the number of years after the present date. Then, 35 + x = 2 (20 + x) Whence, x = — 5, a negative result. On recurring to the problem, we find it so worded as to admit also of the supposition that the epoch is before the pres- ent date; and taking the value of x obtained, with the con- trary sign, we find it will satisfy that enunciation. Hence, a negative result here indicates that a wrong choice was made of two possible suppositions which the problem allowed. From the discussion of these problems we infer : 1. That negatire results indicate cither an erroneous enun- ciation of a 'problem, or a wrong supposition respecting the quality of some quantity belonging to it. 2. That we may form, when attainable, a 'possible problem analogous to that which involved the impassibility, or correct DISCUSSION OF PROBLEMS. 141 the wrong supposition, by attributing to the unknown quan- tity in fin 1 equation a quality directly opposite to that which ltml lice// attributed to it. In general, it is not necessary to form a new equation, l»ut simply to change in the old one the sign of each quantity which is to have its quality changed. Interpret the negative results obtained, and modify the enunciation accordingly, in the following PROBLEMS. 3. If the length of a field be 10 rods, and the breadth 8 rods, what quantity must be added to its breadth so that the con- tents may be 60 square rods ? 4. If 1 be added to the numerator of a certain fraction, its value becomes | ; but if 1 be added to the denominator, it be- comes §. What is the fraction ? 5. The sum of two numbers is 90, and their difference is 120 ; what are the numbers ? 6. A is 50 years old, and B 40 ; required the time when A will be twice as old as B- 7. A and B were in partnership, and A had 3 times as much capital as B. When A had gained $ 2000. and B $ 750, A had twice as much capital as B. What was the capital of each at first ? 8. A man worked 14 days, his son being with him 6 days. and received $39, besides the subsistence of himself and son while at work. At another time lie worked 10 days, and had his son with him 4 days, and received $28. What were the daily wages of each '.' 142 ALGEBRA. XVIII. — ZERO AND INFINITY. 206. A variable quantity, or simply a variable, is a quan- tity to which we may give, in the same discussion, any value within certain limits determined by the nature of the problem ; a constant is a quantity which remains unchanged throughout the same discussion. 207. The limit of a variable quantity is a constant value to which it may be brought as near as we please, but which it can never reach. Thus, if 3 be halved, the quotient § again halved, and so on indefinitely, the limit to which the result may be brought as near as we please, but which it can never reach, is zero. And, in general, if any quantity be indefinitely diminished by di- vision, its limiting value is zero. 208. If any quantity be indefinitely increased by multipli- cation or otherwise, its limiting value is called Infinity, and is denoted by the symbol co . 209. It is evident, from the definition of Art. 207, that if two variable quantities are always equal, their limiting values will be equal. 210. We will now show how to interpret certain forms which may be obtained in the course of mathematical opera- tions. . ' a a Let us consider the fraction - : and let - = a*. b b 1. Interpretation of Let the numerator of remain constant, and the denomi- h nator be indefinitely diminished by division. By Art. l-'>7. it' the denominator is divided by any quantity, the value of the ZERO AND INFINITY. 143 fraction is multiplied by that quantity ; hence the value of the fraction, x, increases indefinitely as b is diminished indefi- nitely. The limiting value of b being (Art. 207), the limit- ing value of - will be - ; and the limiting value of x is co b a (Art. 208). Now - and x being two variable quantities always equal, by xlrt. 209 their limiting values are equal ; or, a a 2. Interpretation of — . 00 Let the numerator remain constant, and the denominator be indefinitely increased by multiplication. By Art. lo8, if the denominator is multiplied by any quantity, the value of the fraction is divided by that quantity ; hence x is diminished indefinitely by division as the denominator increases in- definitely. The limiting value of b being oo, the limiting ft ct value of - will be — : and the limiting value of x is 0. By b co Art. 209 these limiting values are equal ; or, GO Problem of the Couriers. 211. The discussion of the following problem, commonly known as that of Clairaut, will serve to further illustrate the form -, besides furnishing us with an interpretation of the , form pr . Two couriers, A and B, are travelling along the same road, in the same direction, \V R, at the rates of m and n miles per hour respectively. If at any time, say 12 o'clock, A is at the 144 ALGEBKA. point P, and B a miles from him at Q, when and where are they together '.' I ! 1 1 R' P Q B Let #= the required time in hours-, and x = the distance A travels in the time t, or the dis- tance from P to the place of meeting. Then x — a = the distance B travels in the time t, or the dis- tance from Q to the place of meeting. Since the distance equals the rate multiplied by the time, x = m t x — a ==n 1 Solving these equations with reference to t and x, a x = It is proposed now to discuss these values on different sup- positions. 1. in > n. This hypothesis makes the denominator m — n positive; hence the values of both t and x are positive. That is, the couriers are together after 12 o'clock, and to the right of P. This interpretation corresponds with the supposition made. For, if A travels faster than B, he will eventually overtake him, and in advance of their positions at 12 o'clock. 2. in < n. This hypothesis makes the denominator m — n negative; hence the values of both t and x are negative. Now. from what we have observed in regard to negative results < Art. 205), these values of t and x indicate that the couriers w< re together before 12 o'clock, and to the left of P. 111 - - n m a m- - n ZERO AND INFINITY. 145 This interpretation corresponds with the supposition made. For, if A travels more slowly than B, he will never overtake him ; but as they are travelling along the same road, they must have been together before 12 o'clock, and before they could have advanced as far as P. 3. m = n. This hypothesis makes the denominator m — n equal to zero ; ft 77h CL so that the values of t and x become - and -j—, respectively; or, by Art. 210, t = oo and x — co . Since from its nature (Art. 208), go is a value which we can never reach, the values of t and x may be regarded as indicating that the problem is impossible under the assumed hypothesis. This interpretation corresponds with the supposition made. For, if the couriers were a miles apart at 12 o'clock, and were travelling at the same rate, they never had been and never would be together. Thus, infinite results Indicate the imjjossibility of a problem. 4. a = 0, and m > n or m < n. By this hypothesis, the values of t and x each become m — n ' or (Art. 102), £ = and x = 0. That is, the couriers are to- gether at 12 o'clock, at the point P, and at no other time and place. This interpretation corresponds with the supposition made; for, if the distance between them at 12 o'clock is nothing, they are together at P ; but as their rates are unequal, they cannot be together after 12 o'clock, nor could they have been together before that time. 5. a = 0, and m = n. By this hypothesis, the values of t and x each take the form^. 146 ALGEBRA. Referring to the enunciation of the problem, we see that if the couriers were together at 12 o'clock, and were travelling at the same rate, they always had been, and always would be, together. There is, then, no single answer, or finite number of answers, to the problem in this case; and results of this form are therefore called indeterminate. Thus, a result - indicates indeterminathn. 212. The symbol -, however, does not always represent an indeterminate quantity which may have any fin ite mine. Now, in the preceding problem the result - was obtained in conse- quence of two independent suppositions, one causing the nu- merator to become zero, and the other the denominator. We say independent^ because the quantity m — n can be equal to without necessarily causing the quantity a to become 0. And in all similar cases, we should find the result - susceptible of the same interpretation. But if the symbol - is obtained in consequence of the same supposition causing both numerator and denominator to be- come zero, it will be found to have a single definite limiting value. a 2 — b- Take, for example, the fraction — ; if b = a, this single supposition causes both numerator and denominator to become zero, and the fraction takes the form -. Now, dividing both terms by a — b, we have o?-V _ a + b a*-ab~ a ' { ' which equation is true so Ion*; as b is not equal to a. It is not necessarily true when b is equal to a, because the second ZERO AND INFINITY. 147 member was obtained by dividing both terms of the first mem- ber by a — h (which divisor becomes when b = a), as we cannot speak of dividing a quantity by nothing. In (1), as b approaches a, the limiting value of the first member is --, and the limiting value of the second member is 2. Thus we have (Art. 209), Q = 2. Hence the limiting value of the fraction, as b approaches a, is 2. 213. A proper understanding of the theory of indetermi- nation, and of the relation of zero to finite quantities, will lead to the detection of the fallacy in some apparently remarkable results. For example, let a = b Then a i = a b Subtracting //-, a 2 — J 2 = a b — b' 2 Factoring, (a + b) (a — b)—b (a — b) (1) Dividing by a — b, a + b = b (2) But b = a; hence a + a = a then 2 a ■ = a or, 2=1 The error was made in passing from (1) to (2). Equation (1) may be written a + b a — b b a — b Now, as b = a, the second member is an expression of the* form - . But we assumed in going from (1) to (2) that - — - = 1, " ' Cv " (J or that -- = 1 ; which we have seen in Arts. 211 and 212 is not necessarily the case, as it may have any value whatever. 148 ALGEBRA. XIX. — INEQUALITIES. 214. An Inequality is an expression indicating that one of two quantities is greater or less than the other ; as, a > b, and m < n. The quantity on the left of the sign is called the first mem- ber, and that on the right, the second member of the inequality. 215. Two inequalities are said to subsist in the same sense when the first member is the greater or less in both. Thus, a > b, and c > d ; or 3 < 4, and 2 < 3, are inequalities which subsist in the same sense. 216. Two inequalities are said to subsist in a contrary sense, when the first member is the greater in the one, and the second in the other. Thus, a > b, and c < d ; or x < y, and u > z, are inequalities which subsist in a contrary sense. 217. In the discussion of inequalities, the terms greater and less must be taken as having an algebraic meaning. That is, Of" 11 !/ two quantities, a and b, a is the greater when a — b is positive, and a is the less when a — b is negative. Hence, a negative quantity must be considered as less than nothing; and, of two negative quantities, that is the greater which has the least number of units (Art. 49). Thus, > - 2, and - 2 > - 3. 218. An inequality will continue in the same sense after the same quantity has been added to, or subtracted from, each member. INEQUALITIES. 149 For, suppose a > b ; then, by Art. 217, a — b is positive ; consequently, (a + c) — (b + c) and (« — e) — (ft — c) are positive, since each equals a — b. Therefore, a + c > b + c, and a — c > S — c. Hence, it follows that a term may be transposed from one member of an inequality to the other, if its sign be changed. 219. If the signs of all the terms of an inequality be changed, the sign of inequality must be reversed. For, to change all the signs, is equivalent to transposing each term of the first member to the second, and each term of the second member to the first. 220. If two or more inequalities, subsisting in the same sense, be added, member to member, the resulting inequality will also subsist in the same sense. For, let a> b, a'> V, a"> b", then, by Art. 217, a — b, a' — b 1 , a" — b", are all positive ; and consequently their sum a + a' + a" + —b — b' — b"— or, (a + a' + a"+ ) - (b + V + b" + ) i> positive. Hence, a + a'+ a!'+ > b + b' + b" + 221. If two inequalities, subsisting in the same sense, be subtracted, member from member, the resulting inequality will not always subsist in the same sense. 150 ALGEBRA. For, let a > b, and a' > V ; * ■ then a — b and a! — b 1 are positive ; but a — b— (a' — b'), or {a — a') — (b — &')> ma y l )e either positive, negative, or 0. That is, a — a'> b — b', a — a' < & — b', or a — a' = b — V. 222. -'// inequality will continue in the same sense after each member has been multiplied or divided by the same posi- tive quantity. For, suppose a > 5 ; then, since « — J is positive, if m is positive, vi (a — b) and — (a — b) m K ' are positive. That is, m a— m b and are positive. in in Hence, 7 i a b m a > ??i y, and — > — . in m 223. If each member of an inequality he multiplied or di- vided by the same negative quantity, the sign of inequality must be reversed. For, since multiplying or dividing by a negative quantity must change the signs of all the terms, the sign of inequality must be reversed (Art. 219). 224. The solution of an inequality consists in determining the limit in the value of its unknown quantity. This may be done by the application of the preceding prin- ciples. When, however, an inequality and an equation are given, containing two unknown quantities, the process of elimination will be required in the solution. INEQUALITIES. 151 In verifying an inequality, if the symbols of the unknown quantities be taken equal to their respective limits, the ine- quality becomes an equation. EXAMPLES. 225. 1. Find the limit of x in the inequality 23 2x „ i x - -j > -g- + 5. Clearing of fractions, 21 a; — 23 > 2 x + 15 Transposing, and uniting, 19 ic > 38 Whence, x > 2, ^4«s. 2. Find the limits of x in the inequalities, ax + 55x-5«i >a 2 (1) 5x-7ftj; + 7«Ki' 2 (2) from (1), ax + 5b x > a 2 + 5 ab x (a + 5 b) > a (a + 5 b) x > a. From (2), bx — 7ax<b 2 — 7ab x (b - 7 a) < b (b - 7 a) x< b. Hence, x is greater than a, and less than b, Ant;. 3. Find the limits of x and y in the following inequality and equation : 4 x + 6 y > 52 (1) 4 jc + 2 y = 32 (2) Subtracting (2) from (1), 4 ?/ > 20 2/ > 5. (3) From (2), we have y = 16 — 2 x 152 ALGEBRA. Substituting in (3), 16 - 2 x > 5 -2x >-ll 11 -*>-- 2 or (Art. 219), *<T Hence, y > 5, and # 4. Given 5 a? — 6 > 19. Find the limit of cc. 5. Given 2x-5 >25; 3x-7<2x+13. Find tin- limits of x. 6. Given 3 z + 1 > 13 — x; 4cc — 7 < 2 a; + 3. Find an integral value of x. 7. Given 5# + 3?/>46 — y; ?/ — z = — 4. Find the lim- its of x and y. ex c d x cl 8. Given —- + dx — crf> T ; -5 c sc + c d < -^-. Find the limits of x. 9. Given 2x + 3y<57; 2x + y = 32. Find the limits of x and y. 10. A teacher being asked the number of his pupils, replied that twice their number diminished by 7 was greater than 29 ; and that three times their number diminished by 5 was less than twice their number increased by 16. Required the num- ber of his pupils. 11. Three times a certain number, plus 16, is greater than twice that number, plus 24 ; and two fifths of the number, plus 5, is less than 11. Required the number. 12. A shepherd lias a number of slice]) such that three times the number, increased by 2, exceeds twice the number, in- creased by 61; and 5 times the number, diminished by 70, is less than 4 times the number, diminished by 9. How nianv sheep has he ? INVOLUTION. 153 XX. - INVOLUTION. 226. Involution is the process of raising a quantity to any required power. This may he effected, as is evident from the definition of a power (Art. 17), by taking the given quantity as a factor as many times as there are units in the exponent of the required power. 227. If the quantity to he involved is positive, the signs of all its powers will evidently be positive ; but if the quantity is negative, all its even powers will be positive, and all its odd powers negative. Thus, (— a f =(-a)x (— a) X (— a) = + cr X (— a) = — a 3 , (- a y = (- a) X (- a) X (- a) x (- a) = (- a 3 ) X (- a) = + a\ and so on. Hence, Every even poiver is positive, and every odd power has the same sign as its root. INVOLUTION OF MONOMIALS. 228. 1. Let it be required to raise 5 a" b c 3 to the fourth power. 5 a 2 bc 3 x5a 2 bc s x5aHc 3 x5a 2 b c 3 = 625 a 8 b* c v2 , Ans. 2. Raise —3m n 3 to the third power. (— 3 m n 3 ) X (— 3 m n 3 ) X (— 3 m n 3 ) = — 21 m 3 ?j 9 , Ans. RULE. Raise the numerical coefficient to the required power, and multiply the exponent of each letter by the exponent of the re- (j uired power ; making the sign of every wen poiver positive, and the sign of every odd power the same as that of its root. 154 ALGEBRA. EXAMPLES. j Find the values of the following : 3. (a 2 x) 2 . 7. (2x m )\ 11. (-2ab n xf. 4. (-3 cr b) s . 8. (2ab 2 x 8 ) 5 . 12. (-7w*ra) 4 . 5. (-ab 2 c 3 y. 9. (a 2 b 2 )\ 13. (5a 2 6 8 c 4 ) 8 . 6. (a n b) m . 10. (-a 2 c 3 ) 3 . 14. (-6 a; 3 ?/ 7 ) 3 . A fraction is raised to any required power by raising both numerator and denominator to the required power. Thus, 2x*\*_{ 2x*\ ( 2x 2 \ f 2x*\ 8 a; 6 '3f) ~\ 3y) X \ 3y 3 l X \ 3y 3 )~~ 21 if Find the values of the following : 15 . m\ it. L^r. ia i 2x ^ z b J * V 3b I ' V 36 16. R**\\ 18. fL'^V. 20. f-*'^' 4 x y*/ \o / V 4 a~ INVOLUTION OF POLYNOMIALS. 229. Polynomials may he raised to any power, as is obvious from Art. 226, by the process of successive multiplications. Thus, (a + b) 2 =(a + b)(a + b) = a 2 + 2ab + b' 2 , (a + b) 3 =(a + b) (a + b) (a + b) = a 3 + 3 a 2 b + 3 a b 2 + b 3 , and so on. Hence the following RULE. Multiply the polynomial by itself, until it has been taken as a factor as many times as there are units in the exponent of the required power. INVOLUTION. 155 EXAMPLES. Find the values of the following : 1, (a -b) s . 3. {l + a 2 + b 2 )\ 5. (a m -a n )\ 2. (|-^) 2 - 4. (a + m-nf. 6. (a + b) 5 In Chapter XXXVII will be given a method for raising a binomial to any required power, without going through with the process of actual multiplication. SQUARE OF A POLYNOMIAL. 230. It has been shown (Arts. 104 and 105) that the square of any binomial expression can be written down, with- out recourse to formal multiplication, by application of the formulae (a + b) 2 = a 2 + 2ab + b 2 , (a-by = a?-2ab + b 2 . We may also show, by actual multiplication, that (a + b + c) 2 =a 2 + 2ab + 2ac + b 2 + 2bc + c 2 , (a + b + c + d) 2 = a 2 + 2 ab + 2 ac + 2 ad + b 2 + 2 bc+ 2bd + c 2 + 2cd + d 2 , and so on. These residts, for convenience of enunciation, may be writ- ten in another form, (a + b) 2 = a 2 + b 2 + 2ab, (a — b) 2 = a 2 + b 2 — 2 a b, (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2 a c + 2b c, (a + b + c + d) 2 = a 2 + b 2 +c 2 +d 2 +2ab + 2ac + 2ad + 2bc + 2bd + 2cd, and so on. Hence, the following 156 ALGEBRA. RULE. Write the square of each term, together with twice its prod- uct by each of the terms following it. 1. Square x 2 — 2 x — 3. Square of each term, a? 4 + 4 x 2 +9 Twice x 2 X the terms following, — 4. x 3 — 6x 2 Twice — 2 x X the term following, + 12 x » Adding, the result is x A — 4:X 3 — 2x 2 +12x + 9. EXAMPLES. Square the following expressions : 2. a — b + c. 8. 1 + x + x 2 + x 3 . 3. 2x 2 + 3x + 4. 9. x 3 -4x 2 -2x-3. 4. 2x 2 - 3x + i 10. 2x 3 +x 2 +l x-l. 5. a — b — c + d. l\. x 3 + bx 2 — x + 2. 6. »i 3 + 2a; 2 + a;+2. 12. 3x 3 -2 x 2 -x+ ±. 7. 1 — 2 a; + 3 ar. 13. a + & — c — d + e. CUBE OF A BINOMIAL. 231. Hy actual multiplication we may show, (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 , (a-b) 3 = a 3 -3a 2 b + 3a b 2 - b 3 . Hence, for finding the cube of a binomial, the following RULE. Write the cube of the first term, phis three times the square of the first term times the second, 'plus three times the first term times the square of the second, plus the cube of the second term. 3\3 INVOLUTION. 157 EXAMPLES. 1. Find the cube of 2 x 2 - 3 y 3 . (2 a- 2 ) 3 + 3 (2 x 2 ) 2 (- 3 y 3 ) + 3 (2 a: 2 ) (- 3 y 3 ) 2 + (- 3 if) ' = 8z 6 + 3 (4* 4 ) (-3t/ 3 ) + 3 (2 a; 2 ) (9 if) + (-27 y 9 ) _ 8 x G - 36 x* f + 54 x 2 if - 27 y 9 , Jns. Cube the following : 2. a 2 +2b. 4. 3 a; -4. 6. 4 a; 2 -ay. 3. 2m + 5«. 5. 2z 3 -3. 7. 3a;7/ + 5a6 .2 CUBE OF A POLYNOMIAL. 232. By actual multiplication we may show, (a + ft + e y = a 3 + b 3 + c 3 + 3 a 2 b + 3 a 2 c + 3 b 2 a + 3 b 2 c + 3 c 2 a + 3c 2 + Gabc, ( a + l ) + c + d) 3 = a 3 + b 3 + c 3 + d 3 + 3a 2 b + 3a 2 c + 3a 2 d + 3b 2 a + 3 b 2 c + 3b 2 d + 3 c 2 a + 3 c 2 6 + 3 c 2 d + 3 d 2 a + 3d 2 b + 3d 2 c+6abc + 6 a b d + 6 a c d + 6 b c d, and so on. Hence, for finding the cube of a polynomial, the following BULE. Write the cube of each term, tor/ether with three times the product of its square by each of the other terms, and also six times the product of every three different terms. EXAMPLES. 1. Find the cube of 2 a; 2 .— 3 x — 1. 8 x 6 -21 x 3 - 1 - 36 x h - 12 a 4 -f 54 a; 4 - 27 x 2 + Qx 2 -9x + 36x 3 8 x 6 - 36 x s + 42 x 4 + 9 a' 3 - 21 x 2 - 9 x - 1, Arts. 158 ALGEBRA. Find the cubes of the following : 2. a + b — c. 5. 2 - 2 .r + x\ 3. a; 2 -2-1. 6. 1 + a; + x 2 + X 3 . 4. a- 6 + 1. 7. 2.« 3 -r + 2x-3. XXI. — EVOLUTION. 233. Evolution is the process of extracting any required root of a quantity. This may be effected, as is evident from the definition of a root (Art. 17), by determining a quantity which, when raised to the proposed power, will produce the given quantity. It is, therefore, the reverse of involution. 234. Any quantity whose root can be extracted is called a perfect power ; and any quantity whose root cannot be ex- tracted is called an imperfect power. A quantity may be a perfect power of one degree, and not of another. Thus, 8 is a perfect cube, but not a perfect square. 235. To extract any root of a simple quantity, the expo- nent of that quantity must be divided by the index of the root. For, since the ?ith power of a m is a mn (Art. 228), it follows that the nth. root of a mn is a m . 236. Any root of the product of two or more factors is equal to the product of the same root of each of the factors. For, we have seen in Art. 228, in raising a quantity com- posed of factors to any required power, that cadi of the factors is raised to the same power. 237. From the relation of a root to its corresponding power, it follows, from Art. 227, that EVOLUTION. 159 1. The odd roots of any quantity have the same sign as the quantity. Thus, \j a 3 = a ; and ^ — a 5 = — a. 2. The even roots of a 'positive quantity are either positive or negative. For either a positive or negative quantity raised to an even power is positive. Thus, y/ a 4 = a or — a ; or, y' a* = ± a. Note. The sign ±, called the double sign, is prefixed to a quantity ■when we wish to indicate that it is either + or - . 3. Even roots of a negative quantity are not possible. For no quantity raised to an even power can produce a neg- ative result. Such roots are called impossible or imaginary quantities. EVOLUTION OF MONOMIALS. 238. From the principles contained in Arts. 235 to 237, we obtain the following RULE. Extract the required root of the numerical coefficient, ami divide the exponent of each letter by the index of the root; making the sign of every even root of a positive quantity ±, and the sign of every odd root of any quantity the same as that of the quantity. If the given quantity is a fraction, it follows from Art. 228 that we may fake the required root of both of its terms. EXAMPLES. 1. Find the square root of 9 a 4 b- c 6 . \/9a 4 i 2 c 6 = ±3fl 2 i c s , Ans. 2. Find the cube root of - 64 a 9 x z y 6 . .3/ ^ _ 64 a 9 x s y G = — 4:a 3 x y 2 , Ans. 160 ALGEBRA. 8 X 3 7)1 3. Find the cube root of 27 a G b 9 8 / /8 x 3 m 12 \ _ 2 x m 4 Find the values of the following : * 4. if - 125 x 3 y 6 . 9. yV" 1 ^. 14. \/ 729 a 1 * b 24 <A 5. ySla*b 8 . 10. Sl-8an«x\ 15. \/-32 c 6 "^ ™. _ J f 32 m 5 n 10 \ . 5/ V \ 243 J ' 1L ^ 16 a;2m+2 a2 " 16 - V 243 w 16 »*■ 7. \/l»^?. 12. y/ ^QQ^l ) • 17. V(« + *)W- 8. y/ 625 a 12 c 2 . 13. y' 3 2 " 6 3n a n . 18. faj 8B + V 1 " 6 - SQUARE ROOT OF POLYNOMIALS. 239. In Art. 11G we explained a method of extracting the square root of a trinomial, provided it was a perfect square. We will now give a method of extracting the square root of any polynomial which is an exact square. Since the square of a + b is a 2 + 2 a b + b 2 , we know that the square root of a 2 + 2 a b + b 2 is a + b. If we can discover an operation by which we can derive a + b from a 2 + 2 a b + b 2 , we can give a rule for the extraction of the square root. „ „ 7 7 „ 7 Arranmne the terms of the a 2 + 2ab+b 2 a + b 5 °. ' 2 square according to the descend- o „ , 7. 2 a b A- V 1 * n 8 powers of a, we observe thai 2 a b + b 2 the square root of the first term, a 2 , is a, which is the first term of the required root. Subtract its square, a 2 , from the uiven polynomial, and bring down the remainder, 2 a b + b 2 or (2 a + b) b. Dividing the first term of the remainder by 2 a, that is, by twice the first term of the root, we obtain b, the other term. This, added to 2 a, completes the divisor, 2 a + b ; EVOLUTION. 161 which, multiplied by b, and the product, 2 ab + V 2 , subtracted from the remainder, completes the operation. By a similar process, a root consisting of more than two terms may be found from its square. Thus, by Art. 230, we know that (a + b + c) 2 = a 2 + 2 a b + b 2 + 2 a c + 2 b c + c 2 . Hence, the square root of a 2 + 2 a b + b' 2 + 2 a c + 2 b c + c 2 is a + b + c. a 2 + 2ab + b 2 + 2ac+2bc + c 2 •2 a a + b + c 2a + b 2ab + b 2 + 2ac+2bc + c 2 2ab + b 2 2 a + 2 b + c 2ac+2bc + c 2 2ac+2be + r 2 The square root of the first term, a 2 , is a, which is the first term of the required root. Subtracting a 2 from the given poly- nomial, we obtain 2 a b as the first term of the remainder. Dividing this by twice the first term of the root, 2 a, we ob- tain the second term of the root, b, which, added to 2 a, com- pletes the divisor, 2 a + b. Multiplying this divisor by b, and subtracting the product, 2 a b + b' 2 , from the first remainder, we obtain 2 a c as the first term of the next remainder. Doubling the root already found, giving 2 a + 2 b, and di- viding the first term of the second remainder, 2 a c, by the first term of the result, 2 a, we obtain the last term of the root, c. This, added to 2 a + 2 b, completes the divisor, 2 a + 2 b + c ; which, multiplied by the last term of the root, c, and subtracted from the second remainder, leaves no remainder. From these operations we derive the following RULE. Arrange the terms according to the powers of some letter. Find the square root of the first term, write it as the first term of the root, and subtract its square from the given poly- nomial. Divide the first term of the remainder by double the root already found, and add the result to the root, and also to the divisor. 162 ALGEBRA. Multiply the divisor as it now stands by the term of the root last obtained, and subtract the product from the remainder. If there are other terms remaining, continue the operation In the same manner as before. Note. Since all even roots have the double sign ± (Art. 237), all the terms of the result may have their signs changed. In the examples, how- ever, we shall consider only the positive sign of the result. EXAMPLES. 1. Find the square root of 9 x* — 12 x s + 16 x 2 — 8 x + 4. 9a; 4 -12a; 3 + lGx~-8x + 4: 3a; 2 -2a; + 2 9 a; 4 6 a: 2 - 2 ./' -12 a; 3 - 12 x 3 4 a;' 2 6 x- - 4 x + 2 12 x 2 - 8 x + 4 12 a;' 2 -. 8 x + 4 Ans. 3x 2 — 2x + 2. Find the square roots of the following : 2. ±x i -±x s -3x 2 + 2x + l. 2 1 — + — i m m 3. 4 a 4 -16 a 3 +24 a 2 - 16 a + 4. 4. m 2 + 2 m — 1 5. 9 — 12 x + 10 :•■- - 4 x 3 + x*. 6. 19 x 2 + 6 x 3 + 25 + x i + 30 a-. 7. 28 a- 3 + 4 a; 4 - 14 x + 1 + 45 x 2 . 8. 40 jc + 25 - 14 x 2 + 9 x* - 24 a; 3 . 9. 4 .r 4 + 64 - 20 a- 3 - 80 x + 57 x 2 . 10. a 2 + b 2 + c 2 -2 ab -2 a c + 2 b c 11. a; 2 + 4 y 2 + 9 « 2 — 4 a; ?/ + 6 a; » — 12 y z. No rational binomial is an exact square;; hut, by the rule, the (ip/imxliiinte root may be found. EVOLUTION. 10- • i Find, to four terms, the approximate square roots of the fol- lowing : 12. 1 + x. 13. a 2 + b. 14. 1 — 2 x. 15. a 2 + x 2 . The square root of a perfect trinomial square may be ob- tained by the rule of Art. 116, Find the square roots of the first and last terms, and con- nect the results by the sign of the second term. Extract the square roots of the following : 16. x 4 + 8x 2 +16. 19. « 2m -4 a m+n + 4 a 2 ". rtrt a 2 4 a ±b 2 17. 9x*-6xf + f. 20. _- — +^. t 2 4 x 2 9 ?/ 4 18. a*-ax + T . 8L 9? + 2 +4^- SQUARE ROOT OF NUMBERS. 240. The method of Art. 239 may be used to extract the square roots of numbers. The square root of 100 is 10 ; of 10000 is 100 ; of 1000000, is 1000 ; and so on. Hence, the square root of a number less than 100 is less than 10 ; the square root of a number between 10000 and 100 is between 100 and 10 ; the square root of a number between 1000000 and 10000 is between 1000 and 100 ; and so on. Or, in other words, the integral part of the square root of a number of one or two figures, contains one figure; of a number of three or four figures, contains two figures ; of a number of five or six figures, contains three figures ; and so on. Hence, If a point is placed over evei*y second figure in any integral number, beginning with the units' 1 place, the number of point* will shoiv the number of figures in the integral part of its square root. 164 ALGEBRA. 241. Let it be required to find the square root of 4356. Pointing the number according to 60+6 the rule of Art. 240, it appears that there are two figures in the integral 4356 3600 120 + 6 756 'J" ° ~~— -~ . jgQ part oi the square root. Let a denote the figure in the tens' place in the root, and b that in the units' place. Then a must be the greatest multiple of 10 whose square is less than 4356 ; this we find to be 60. Subtracting a 2 , that is, the square of 60, or 3600, from the given number, we have the remainder 756. Dividing this remainder by 2 a, or 120, gives 6, which is the value of b. Adding this to 120, multiplying the result by 6, and subtracting the product, 756, there is no remainder. Therefore we conclude that 60 + 6, or 66, is the required square root. The zeros being omitted for the sake of brevity, we may ar- range the work in the following form : 4356 36 G6 126 756 756 RULE. Separate the given number into periods, by pointing every second figure, beginning with the units' [dace. Find the greatest square in the left-hand period, and place its root on the right ; subtract the square of this root from the first period, and to the remainder bring down the next period for a dlr hi end. Divide this dividend, omitting the last figure, by double the root already found, and annex the result to the root and also to the divisor, Multiply tin' divisorj as it now stands, by the figure of the root last obtained, and subtract the product from tin- dividend. If there are more periods to be brought down, continue the operation in the same manner as before. EVOLUTION. 165 If there be a final remainder, the given number has not an exact square root ; and, since the rule applies equally to deci- mals, we may continue the operation, by annexing periods of zeros to the given number, and thus obtain a decimal part to be added to the integral part already found. It will be observed that decimals require to be pointed to the right ; and if they have no exact root, we may continue to form periods of zeros, and obtain decimal figures in the root to any desirable extent. As the trial divisor is necessarily an incomplete divisor, it is sometimes found that after completion it gives a product' larger than the dividend. In such a case, the last root figure is too large, and one less must be substituted for it. The root of a common fraction may be obtained, as in Art. 238, by taking the root of both numerator and denominator, when they are perfect squares. If the denominator only is a perfect square, take the approximate square root of the nu- merator, and divide it by the square root of the denominator. If the denominator is not a perfect square, either reduce the fraction to an equivalent fraction whose denominator is a per- fect square, or reduce the fraction to a decimal. EXAMPLES. 1. Extract the square root of 49.434961. 49.434961 49 7.031 1403 4349 4209 14061 14061 14061 Ans. 7.031. Here it will be observed that, in consequence of the zero in the root, we annex one zero to the trial divisor, 14, and bring down to the corresponding dividend another period. Extract the square roots of the following : ALGEBRA. 6. .9409. 10. .006889. 7. 6561 9025' 11. .0000107584 8. 1.170724. 12. 811440.64. 9. 446.0544. 13. .17015625. 166 2. 273529. 3. 45796. 4. 106929. 5. 33.1776. Extract the square roots of the following to the fifth decimal place : 14. 2. 16. 31. 18. 15. 5. 17. 173. 19. 242. When n + 1 figures of a square root have been ob- tained by the ordinary method, n more may be obtained by simple division only, supposing 2n + l to be the whole num- ber. Let N represent the number whose square root is required, a the part of the root already obtained, x the rest of the root ; then y/_ZV= a + x, whence, iV= a 2 + 2 a x + x 2 ; therefore, JV — a 2 = 2 a x + x 2 , 7 1 20. 9' 3' 3 2 21. 16' N-a* = x + x~ 2 a '2a Then iV— a 2 divided by 2 a will give the rest of the square x root required, or x, increased by -= — ; and we shall show that x -= — is a proper fraction, less than \, so that by neglecting the Zi a remainder arising from the division, we obtain the part re- quired. For, x by supposition contains n figures, so that x 2 cannot contain more than 2 n figures ; but a contains 2n + l EVOLUTION. 167 figures ; and hence — is a proper fraction. Therefore - a 2 a is a proper fraction, and less than ^. In the demonstration we supposed JV an integer with an exact square root ; but the result may be extended to other cases. From the examples in Art. 241, we observe that each re- mainder brought down is the given expression minus the square of the root already obtained ; and is therefore in the form A 7 " — a 2 . If, then, any remainder be divided by twice the root already found, we can obtain by the division as many more figures of the root as we already have, less one. We will apply these principles to calculating the square root of 12 to the. sixth decimal place. We will obtain the first four figures of the result by the ordinary method : 12.000000 9 3.464 64 300 256 686 4400 4116 6024 28400 27696 '04 The remainder now is .000704 ; and twice the root already found is 6.928. Then, by dividing .000704 by 6.928, we can obtain the next three figures of the root. Thus, 6.928). 0007040 (.000102 .0006928 11200 That is, the square root of 12 to the nearest sixth decimal place is 3.464102. The following rule will be found to save trouble in obtaining approximate square roots by this method : 168 ALGEBRA. Divide the remainder by twice the root already found {omit- ting the decimal point), and annex all of the quotient, except the decimal point, to the part of the root already found. In practice the work would be arranged thus : 12. 9 64 30C 25C 3.464 ) 686 4400 4116 6924 : 28400 27696 6928) 704.000 (.102 6928 11200 Am. 3.464102 EXAMPLES. 1. Extract the square root of 11 to the 4th decimal place. 2. Extract the square root of 3 to the 6th decimal place. 3. Extract the square root of 61 to the 8th decimal place. 4. Extract the square root of 131 to the 3d decimal place. 5. Extract the square root of 781 to the 5th decimal place. 6. Extract the square root of 12933 to the 4th decimal place. CUBE ROOT OF POLYNOMIALS. 243. Since (a + b) 3 = a s + 3 a 2 b + 3 a b 2 + b 3 , we know that the cube root of a 3 + 3 a" b + 3 a b 2 + b 3 is a + b. a 3 + 3 a 2 b + 3 a b 2 + b 3 a" a + b 3a 2 +3ab + b 2 3 a 2 b + 3 a b 2 + b* 3a 2 b + 3ab 2 + b 5 EVOLUTION. 1G9 Arranging the terms of the cube according to the descending powers of a, we observe that the cube root of the first term, a 3 , is a, which is the first term of the required root. Subtract its cube, a 3 , from the given polynomial, and bring down the re- mainder, 3 a 2 b + 3 a b' 2 + b 3 or (3 a 2 + 3 a b + b' 2 ) b. Dividing the first term of the remainder by 3 a 2 , that is, by three times the square of the first term of the root, we obtain b, the other term of the root. Adding to the trial divisor 3 a b, that is, three times the product of the first term of the root by the last, and b' 2 , that is, the square of the last term of the root, completes the divisor, 3 a' 2 + 3 a b + b' 2 ; which, multiplied by b, and the product, 3 a 2 b + 3 a b' 2 + b 3 , subtracted from the remainder, completes the operation. If there were more terms, we should proceed with a + b exactly as previously with a ; regarding it as one term, and dividing the first term of the remainder by three times its square ; and so on. Hence, the following RULE. Arrange the terms according to the powers of some letter. Find the cube root of the first term, write It as the first term of the root, and subtract its cube from the given polynomial. Take three times the square of the root already found for a trial divisor, divide the first term of the remainder by it, and write the quotient for the next term of the root. Add to the trial divisor three times the prod net of the first term by the second, and the square of the second term. Multiply the complete divisor by the second term of the root, and subtract the product from the remainder. If there are other terms remaining, consider the root already found as one term, and proceed as before. EXAMPLES. 1. Find the cube root of x 6 — 6 x'° + 40 x 3 — 96 x — 64. 170 X X ALGEBRA. 6 -6a 5 +40a; 3 -96a-64 1 x 2 -2j 6 3x i -6x a +4:x i —6 a- 5 — 6 a; 5 +12 a 4 - 8 x s 3 a 4 - 12 a 3 + 12 a; 2 -12a 2 +24a; + 16 -12a; 4 +48a; 3 3a 4 -12a: 3 + 24a + 16 - 12 x* +48 x 3 - 96 .r -64 Ans. x' 2 — 2 a; — 4. The formation of the second divisor may be explained thus : Regarding the root already obtained, a 2 — 2 a, as one term, three times its square gives 3 a 4 — 12 a 3 + 12 x' 2 ; three times x~ — 2 a; times — 4 ; gives — 12 x' 2 + 24 x ; and the square of the last root term is 16. Adding these results, we have for the complete divisor, 3 a 4 — 12 x 3 + 24 x + 16. Find the cube roots of the following: 2. 1 — 6 y + 12 y' 2 - 8 ?/ 3 . 3. 8 a; 6 + 36 a; 4 + 54 a; 2 + 27. 4. 64 a; 3 - 144 a b x 2 + 108 a 2 i 2 a: - 27 a 3 6 3 . 5. a- 6 + 6 a- 5 - 40 a 3 + 96 x — 64. 6. v/«-l + 57/ 3 -37/ 5 -3y. 7. a: 3 + 3a;H h-g. 8. 15 r 4 - 6 a- - 6 a; 5 + 15 x 2 + 1 + a 6 - 20 x*. 9. a a + 3 a a & + 3a 2 c + 3aJ 2 + 6a6c + 3ac 2 +S 8 + 3 6 a c + 3l> r 2 + c 8 . 10. 9 a 3 - 21 x 2 - 36 a 5 + 8 a; 6 - 9 x + 42 a- 4 - 1. No rational binomial is an exact cube; but, by the rule, the approximate root may be found. EVOLUTION. 171 Find, to four terms, the approximate cube roots of the following : 11. X s + 1. 12. x s — a° 13. 8 ./• o CUBE ROOT OF NUMBERS. 244. The method of Art. 243 may he used to extract the cube roots of numbers. The cube root of 1000 is 10; of 1000000, is 100; of 1000000000, is 1000; and so on. Hence, the cube root of a number less than 1000 is less than 10 ; the cube root of a num- ber between 1000000 and 1000 is between 100 and 10 ; the cube, root of a number between 1000000000 and 1000000 is betw T een 1000 and 100 ; and so on. Or, in other words, the integral part of the cube root of a number of one, two, or three figures, contains one figure; of a number of four, five, or six figures, contains two figures ; of a number of seven, eight, or nine figures, contains three figures ; and so on. Hence, If a point is placed over every third figure in any integral a miller, beginning with the units' place, the number of points will show the number of figures lit the integral part of its eube root. 245. Let it be required to find the cube root of 405224. Pointing the number according to the rule of Art. 244. it appears that there are two figures in the integral part of the cube root. Let '/ denote the figure in the tens' place in the root, and b that in the units' place. 405224 343000 70 + 4 14700 840 16 15556 62224 62224 Then a must be the greatest mul- tiple of 10 whose cube is less than 405224 ; this we find to be 70. Subtracting a 3 , that is, the cube of 70, or 343000, from the given number, we have the remainder 62224. Divid- ing this remainder by 3 a' 2 , or 14700, gives 4, which is the 172 ALGEBRA- value of b. Adding to the trial divisor 3 a b, which is 840, and b' 2 , which is 16, completes the divisor, 15556. Multiplying the result by 4, and subtracting the product, 02224, there is no remainder. Therefore we conclude that 70 + 4, or 74, is the required cube root. The work is usually arranged thus : 405224 343 74 14700 840 16 15556 62224 62224 RULE. Separate the given member into periods, by pointing every third figure, beginning ivith the units' place. Find the greatest cube in the left-hand period, and place its root on the right / subtract the cube of this root from the left- hand period, and to the remainder bring down the next period for a dividend. Divide this dividend, omitting the last two figures, by three times the square of the root already found, and annex the quo- tient to the root. Add together the trial divisor, with two zeros annexed; three times the product of the last root figure by the rest of the root, ivith one zero annexed ; and the square of the last root figure. Multiply the divisor, as it now stands, by the figure of the root last obtained, and subtract the product from the dividend. If there are more periods to be brought down, continue the operation in the same manner as before, regarding the root already obtained as one term. The observations made after the rule for the extraction of the square root (Art. 241) are equally applicable to the extrac- tion of the cube root. EVOLUTION. 173 EXAMPLES. 1 Extract the cube root of 8.144865728. 8.144865728 8 2.012 120000 600 1 120601 144865 120601 12120300 12060 4 24264728 24264728 12132364 Ans. 2.012. Here it will be observed that, in consequence of tbe in the root, we annex two additional zeros to the trial divisor, 1200, and bring down to tbe corresponding dividend another period. Extract the cube roots of the following : 2. 1860867. 4. 1481.544. 6. 51.478848. 3. .724150792. 29791 681472 7. .000517781627. Extract the cube roots of the following to the third decimal place : 8. 3. 10. 212. 12 4 9. 7. 11. 5 8 13 3 13 ' 17 246. JVJien n + 2 figures of a cube root have been obtained by the ordinary method, n more may be obtained by division only, supposing 2 n + 2 to be the whole number. 174 ALGEBRA. Let N represent the number whose cube root is required, a the part of the root already obtained, x the rest of the root ; then $ 2V= a + x, whence, iV= a 3 + 3 a 2 x + 3 a x 2 + x s ; therefore, _A r — a 3 = 3 a 2 x + 3 a x 2 + x 3 , iV" — a 3 x 2 x 3 » + — + dec a 6 a- Then A 7 "— a 3 divided by 3 a 2 will give the rest of the cube root required, or x, increased by - 1- 77—,; and we shall show a o a that the latter is a proper fraction, less than h, so that by neglecting the remainder arising from the division, we obtain the part required. For, x by supposition contains n figures, so that x 2 cannot contain more than 2 n figures. But « con- o X" tains 2 n + 2 figures ; and hence - — is less than ^ . And as n — 5= — X o — , and - — is less than 1, - — . must also be less o cr a 6 a 6 a -6 a 1 than T T n . Therefore, (- ^ — ^ is a proper fraction, less than \. a k> a Remarks similar to those in the last part of Art. 242 apply here. ANY ROOT OF POLYNOMIALS. 247. In order to establish a general rule for the extraction of roots, it will be necessary to notice the formation of the n\\\ power of a polynomial, n being any entire number whatever. Thus, (a + by = a n + n a n ~ } b + Therefore, y' a n + n a u ~ l b + = a + b. The first term of the root, a, is the nth root of a n , the first term of the power; and the .second term of the root, b, may be EVOLUTION. 175 obtained by dividing the second term of the power by n a n ~\ or by n times the (n — l)th power of the first term of the root. If the root now found be raised to the nth power, and sub- tracted from the given polynomial, it will be seen that two terms of the required root have been determined. It will be observed that the process is essentially that of the preceding Articles, simplified by dispensing with completed divisors, and generalized. Hence the • RULE. Arrange the terms according to the powers of some letter. Find the required root of the first term for the first term of the root, and subtract its power from the given polynomial. Divide the first term of the remainder by n times the (n — l)th power of this root, for the second term of the root, and subtract the nth power of the root now found from the given polynomial. If other terms of the root require to be determined, use the same divisor as before, and proceed in like manner till the nth poiver of the root becomes equal to the given polynomial. This rule is, also, applicable to numbers, by taking n figures in each period. EXAMPLES. 1. Extract tl* cube root of x 6 + 6 x 5 — 40 x s + 96 x — 64. x 6 + 6 x 5 - 40 x 3 + 96 x - 64 (* 2 ) 2\3 x 6 x 2 + 2x 3 a; 4 1 6 x 5 (x 2 + 2 x) 3 = x 6 +6 x 5 + 12x*+Ssci a Sx 4 -12* 4 (x 2 + 2x-4) 3 = x 6 + 6 x 5 - 40 x 3 + 96 x - 64 Ans. x- + 2 x — 4. 2. Extract the cube root of ??i 6 — 6 m 5 + 40 m 3 — 90 m — 64. 3. Extract the square root of a i —2a 3 x + 3a 2 x 2 —2ax 3 + x i . 176 ALGEBRA. 4. Extract the fifth root of 32 x 5 - 80 x* + SO x s — 40 x" + 10 a; — 1. 5. Extract the fourth root of x s — 4 a 7 + 10 a; 6 — 16a; 5 + 19 x 4 - 16 a- 3 + 10 x 1 - 4 x + 1. 248. When the index of the required root is a multiple of two or more numbers, we may obtain the root by successive extractions of the simpler roots. For, since (Art. 17), ( 7 a) mn = a, taking the nth. root of both members, we have (Art. 235), Taking the mth root of both members, y/ a== y (y a). Or, the mnth root of a quantity is equal to the mth root of the nth root of that quantity. EXAMPLES. 1. Extract the fourth root of 16 x i - 96 x 3 y + 216 x 2 y 2 - 216 x y s + 81 y\ 2. Extract the sixth root of a 12 — 6 a 10 + 15 a 8 — 20 a 6 + 15 a 4 - 6 a- + 1. 3. Extract the fourth root of m 8 — 8 m 1 ± 12 w 6 + 40 m 5 - 74 m* - 120 m s + 108 m 2 + 216 wi + 81. XXII. — THE THEORY OF EXPONENTS. 249. In Art. 17, we defined an exponent as indicating how many times a quantity was taken as a factor; thus a m means ay. ay, a to m factors. Obviously this definition has no meaning unless the expo- nent is a positive integer; and as fractional and negative ox- EXPONENTS. 177 portents have not been previously defined, we may give to them any definition we please. 250. We found (Arts. 82, 93, and 228) that when m and n were positive integers, I. a m Xa n — a m + n . a m II. — = a m ~ n . a n III. («'")" = a mn . As it is convenient to have all exponents follow the same laws, as regards multiplication, division, and involution, we shall define fractional and negative exponents in such a way as to make Ride I hold for any values of m and n. We shall now find what meanings must be assigned to them in con- sequence. 3 251. To find the meaning of a?. As Rule I is to hold universalh T , it follows that a a a + a s a 2 Xa 2 = a 2 2 = a 2 =a 3 . a Hence, a 2 is such a quantity as when multiplied by itself 3 produces a 3 . Then, by the definition of root (Art. 17), a 2 must a be the square root of a? ; or, a 2 = \J a z . Again, to find the meaning of a 3 . i i i. i + i + i a By Rule I, a 3 X a, 3 X a 3 ■= a 3 3 3 = a 3 = a. Hence, a* is such a quantity as when taken 3 times as a factor produces a ; or, a 3 = fya. 252. We will now consider the general case. p To find the meaning of a q > p and a being positive integers. 178 ALGEBRA. p_ p p By Kule I, a q X a 9 X a 5 X to q factors P P P P yy — +—-i 1- to? terms — X<2 = a q q « =a q =a p . p Hence, a 5 is such a quantity as when taken q times as a p factor produces a p . Then a' 1 must be the qi\\ root of a p ; p or, a* = ya p . 3 4 5 1 For example, a* = \j a z \ c 2 =\J c 5 ; x 3 ' = y x ; etc., and, conversely, y' a 5 = « 4 ; y 7 ic = x 2 ; y" m 5 = m, 3 " ; etc. EXAMPLES. 253. Express the following with radical signs instead of fractional exponents : a 2 x i a 3.2 c 2 . 5. x*y^. 7. 4a 5 J« 9. 5y T °£ T ^. 2.6^. 4. 3«m^. 6, rn'iA 8. 2c«t/l 10. 3»^c^l Express the following with fractional exponents instead of radical signs : 11. yV. 13. sjn. 15. 3 y/ m 5 . 17. ty a 1 ty a?. 12. yV- 14. yV. 16. 4 ^a 10 . 18. v^vV- 19. 5sjm»%n<. 20. 2 av Vy>. 254. To find the meaning of a -3 . By Kule I, a~ 3 X a 3 = a = 1, by Art. 94. Hence, a ~ 3 = — - . To find the meaning of a~ 2 . By Rule I, a~* X c$ — a = 1. Hence, a. 2 = — . EXPONENTS. 179 255. "We will now consider the general case. To find the meaning of a~ s , s being integral or fractional. By Eule I, a~ s X a s = a° = 1. 1 Hence, a s = — . 1 1 -2 1 For example, ar 1 = ~\ a 4 — ^?5 a 3 — ~ '■> e * c v a' a a* 1 X 2 2-3 and, conversely, — ^ = « -2 ; — = a; 2 a -3 ; — = 2 a * ; etc. We observe, in this connection, the following important principle : ^4 quantity may be changed from the denominator of a frac- tion to the numerator, or from the numerator to the denomi- nator, if the sign of its exponent be changed. EXAMPLES. 256. Remove all powers from the denominators to the numerators in the following : ar 5a; 3 2 a; -1 a; a; 2 , a; -2 x~ a 2 a 3 — 1 a 4 a 5 — b 3. 4. x 3 3 a- 4 + a; 2 a; 5 7 m 6c- 1 3 m 3 7 c* 4m 2 -l 3??? 3 + 2?i i + 5 c 6 2 c" 5 Remove all powers from the numerators to the denominators in the following : *o _ 2a; 3a" 2 a _ a? a: 3 x~ 2 2x 1 180 ALGEBRA. a 6 3a 4 5a 2 a 7. —^--^ — ^ + 8. a; + 2 5 b 03 7 — b a' m — x m 3 % 5 2p 1 — x 2 3x 5x 1 7 x K ™ — 1 7 ^. — 3 * Express the following with positive exponents: J_ 2. _3 9. 2x 2 y 2 — 3x~ l y i '— x~ i y r . 10. a- 1 5- 2 + 2cr 3 i- 4 -3fth~l _i _s _i 11. 3a; 3 y 7 — 4 a* ?/ + a y~ • 12. a" 1 6~ 2 c 8 + a~ 2 b~% e -4 + a s j-2 c# 257. We obtained the meanings of fractional and negative exponents on the supposition that Bule I, Art. 250, was to hold universally. Hence, for any values of m and n, a m x a n — a m + n 3 2 3_2. J^ For example, a 2 X «~ 5 = a 2 ~ 5 = a~ 3 ; a 4 X « — «* 3 = a ' 5 a- i Xa* = a 2 = a 2 ; a 3 X a 5 = a 3 ° = a To I etc. EXAMPLES. Multiply together the following : 1. a 8 and a" 1 . 4. c 8 and y'c 2 . 7. rc and n~?. 2. a 2 and a" 2 . 5. a:" 1 and (far 8 . 8. a^ and af*. 3. a- x and a~ 5 . 6. m 2 and ^— . 9. 2 c~ * and - 3 a tyc s . EXPONENTS. 181 10. Multiply A ^+2a*-3ftHy26 *-4a 3_6a £&*. a% b~ 2 + 2 a^ - 3 b 2 2 J~2 _ 4 a -J _ o a ~% iih 2 a$ &- 1 + 4 a* 6"^ - 6 Aa*b~*-S + 12a H 2 . 6 - 12 a - J b 2 + 18 a %b 2 a$ b- 1 - 20 + 18 a » 6, ^ws. Note. It should be carefully remembered, in performing examples like tbe above, that any quantity whose exponent is is equal to 1 (Art. 94). Multiply together the following : 11. a 2 b~°- - 2 + a~ 2 b 2 and a 2 b~ 2 + 2 + a~ 2 b 2 . 3. i i ii a , i i 12. a 4 — a? b± + a 4 b' 2 - b* and a 4 + b*. 13. a~ 2 - 2 a" 1 b + b 2 - a b 3 and ar s + 2 or 2 b. 14. 3 ar 1 -a~ 2 b- 1 + or* b~ 2 and 6 a* b 2 + 2 a 2 b + 2a. 15. x~ s f -x~ 2 y-2x- 1 and 2 x 2 y~ l + 2x 8 ! /- 2 -±x i y~\ 16. x% y~* + 2 + jc" * y* and 2 a; - * y 4 _ 4 A .- $ y f + 2 a-- 2 y 4 . 17. 2 sc^ — 3 a;^ — 4 + aT* and 3 a;'*' + x — 2 x*. 18. 4 a 4 &- 1 + « 4 - 3 «~ 4 6 and 8 a 4 &- 1 - 2 a~± - 6 «~ 4 6. 258. To prove that Rule II holds for all values of 'm and n. By Rule I, a m ~ n X a n = a m ~ n + w = a m . Inverting the equation, and dividing by a n , we have — = a" a n 182 ALGEBRA. ft 3 cc~ 2 For example, — =:a 3_1 =:a 2 ; — — = a~ 2 ~ 3 = a~ 5 ; -I _| + 2 f a 3 8 + 4 V- a - = a 4 ' " = a 4 ; — - = a 5 = a 5 ; etc. EXAMPLES. Divide the following: _i _4 l 1 1. a 3 by a -1 . 4. a 2 by a . 7. x by -^— g. 2. a by a 3 . 5. rMiyJc 5 . 8. 5 » by 2 or 1 $ 6. 3. a^ by a* 6. m 2 by tym,- 2 . 9. a" 1 6^ by - 3 a 6~~*. 10. Divide 2 a^ 6" 1 -20 + 18 a - """ 6 by J b~ * + 2 J - 3 6*. 2 J' b- 1 - 20 + IS a ^b 2a?b~ 1 +4 : cfib~^—6 2 _l JL I a :i 6 »+-2a* — 3 6' 2 6 2 - 4 a * — 6 a $ b 2 , ^?*s. -4 a* 6 * — 14 + 18 a ^6 -4a^6~^- 8 + 12 »~^ 6^ -6 -12 a"* 6^+ 18 a~ %b - 6 - 12 a~^ 62 + 18 <T% b Note 1. Particular attention must be given to seeing that the dividend and divisor are arranged in the same order of powers, and that each re- mainder is brought down in the same order. It must be remembered that a zero exponent is greater than any negative exponent ; and that negative exponents are the smaller, the greater their absolute value. Note 2. In dividing the first term of the dividend or remainder by the first term of the divisor, it will be found convenient to write the quotient at first in a fractional form; reducing the result by the principles of Art. 258. Thus, in getting the first term of the quotient in Ex. 10, we divide 2 a* b- 1 by a? b ' 2 . Then, the result = „ _ L = 2 a 3 3 b + * = 2 b * (fit * EXPONENTS. 183 Divide the following : jl i 11. a — b by a 5 — b b . 12. a- 4 + a~ 2 b' 2 + b~ i by a~ 2 + a' 1 b' 1 + b~ 2 . 13. 2x- 2 y 2 + 6 + Sx 2 i/- 2 hj2x + 2x 2 y- 1 + 4 t x 3 y- 2 . 14. 2 x ?s y~ x — 2x~%y + 32 x~ 2 y 3 by 2 + 6 a-" § y + 8 as" * ?/. 15. cc~ 3 ?/ -5 — 3 x~ 5 y~ n + x -1 y~ 9 by x~ 2 y~ s + x~ 3 ?/ -4 — a; - 4 y -6 . 16. 8 — 10o;- 2 2/ J ^+2aj-*2/^ > " by 4x~*y% + 2x~ 2 y*—2x~^y 4 . 259. To prove that Rule III holds for all values of m and n. We will consider three cases. Case I. Let m have any value, and n be a positive in- teger. Then, from the definition of a positive integral exponent, (a m ) n = a m Xa m Xa m to n factors — f,m + m + m ton terma __ ~m n Case II. Let m have any value, and n be a positive frac- v tion, which we will denote bv - • P_ q, Then, (a m ) n = (a m )T= \ (a m ) p , by the definition of Art. 252, = y 7 ^"^ by Case I, Art. 259, mp = aJ, by Art, 252, = a mx q = a mn . Case III. Let m have any value, and n be a negative quantity, integral or fractional, which we will denote by — s. 184 ALGEBRA. Then, (a m ) n = (a m )~ s — — -r s , by the definition of Art. 255, (a m ) s = , by Cases I and II, Art. 259, = a - ms = a m{ - s) = a mn . Thus, we have proved Kule III to hold for all values of m and n. For example, (a 2 ) 3 = a 6 ; (a" 1 ) 5 = a" 5 ; (a~ ?s ) * = a"* ; (J)$ = a; (a*)~* = a ~*J (a 2 )~* = a-*; etc. EXAMPLES. 260. Find the values of the following : 5 '( C " *)">. 10. f ] 1. («•)-■ 4. (O - *- 7. tf( c 2 ) 2 . 10. - . 2. («- 2 ) 3 . 5. (e-*)- 2 *. 8. tfm 8 ) * 11. (^ 3. (a 3 )*. 6. (,/*)* 9. (^f)- 5 . 12. {(^VY 1 - 261. To prove that (a £) n = a n b n for any value of n. In Art. 228 we showed the truth of the theorem when n was a positive integer. Case I. Let n be a positive fraction, which we will denote v E E E by — . We have then to show that (ab)« =a« b'i. 9. [(a bfy = (a b) p , by Art. 259. p p EXPONENTS. 185 [at &*]* = (c£y (b^) q = a p bP= (ab)", by Art. 228. Hence, [( a J)fj*=[a? J?] 3 . Therefore, (a fi) «= a« &«. Case II. Let w be a negative quantity, which we will de- note by — s. We have then to show that (a b)~ s = a~ 3 b~ 3 . (a b)-° = -y—^- = — ; - , by Art. 228 and Case I, v J (a b) s a s b s J = ar s b~ s . 262. To find the value of a numerical quantity affected with a fractional exponent. 1. Find the value of 8*. From Art. 252, we should have S^^y'S 5 ; and to find the value in this way, we should raise 8 to the fifth power, and take the cube root of the result. A better method, however, is as follows : 8^ = (8*) 5 , by Art. 259, = ($&)* = 2 s = 32; Ans. Note. Place the numerator of the fractional exponent as the exponent of the parenthesis, and 1 divided by the denominator as the exponent of the cpiantity within. 2. Find the value of 16"*. | = J__ J_ J_ J_ 1_ ~ 16* ~ (IB*)* = ^ 16 ) 5 ~ (± 2 > 5 ~ ±32 : 186 ALGEBRA. EXAMPLES. Find the values of the following : 3.27?. 5.1000-* 7. (-8)1. 9. ('f^f . 36' 2 x 16 * 1 7 s 4^ x 9~ 2 4. 36*. 6. 9"* 8. (-27)* 10. -^T 1 81" X 16* If the numerical quantity is not a perfect power of the de- gree indicated by the denominator of the fractional exponent, the first method explained in Ex. 1, Art. 262, is the best. For example, to find the value of 7 2 , we write it \/ 7 s , or y/ 343 ; and find the square root of 343 to any desired degree of accuracy. MISCELLANEOUS EXAMPLES. 263. Extract the square roots of the following : 4 x y c*de 2 5. 9 x-* y 2 -12x~ s y-2 x~ 2 + 4 x~ x y~ x + y~\ 6. 4x* + ±x% y~* -15x 2 y~* -Sx% y~* + lGx§ y~\ 7. z 8 y~% + 6 - 4 x~% y% + x~ 3 y% - 4 x% y~*. V Extract the cube roots of the following : 8. ab\ 9. -8a;- 4 / 10. 11. $ if - 12 y^'x-^ 6 y$x- 2 -i/x- 3 . 3m 2 n * ax 5 EXPONENTS. 187 Reduce the following to their simplest forms : 12. ,^„„^. . 15. a x ~ v+2z a 2x+v ~ 3z a z . Ji + 2»i+r 13. (x a )- b +(x- a )- b . 16. (^'"x^X^- 1 ) *. ia (a x+v Y ( a? \ x ~ v ,_ r/ _J_\„_»-i_2_ Change the following to the form of entire quantities : 18 15 ^* 2 19 X *V 2 20 ^ 2 Reduce the following to their simplest forms : 21. ' ^ 22. ^=^. 23 Factor the following expressions : 24. 9^-12^ + 4. 25. a ^-3a*-88. 26. ar 2 & + 5 a" 1 &*- 66. Factor by the method of Art. 117 : 27. a-b. 28. a£-&~£. 29. ar'"y — 4«*. Factor by the method of Art. 119 : 30. a — 6. 31. a + 6. 32. x~ s +8cm^. 188 ALGEBRA. XXIII. — RADICALS. 264. A Radical is a root of a quantity, indicated by a radical sign ; as, yfa, \Jx + 1, y m 2 — 2 n + 3. When the root indicated can be exactly obtained, it is called a rational quantity ; and when it cannot be exactly obtained, it is called an irrational or surd quantity. 265. The Degree of a radical is denoted by the index of © / the radical sign; thus, \] a is of the second degree; \x + 1 of the third degree. Similar Radicals are those of the same degree, with the 5/- 6, same quantity under the radical sign ; as, \ax and 1 y ax. 266. Most problems in radicals depend for their solution on the following important principle : For any values of n, a, and b, by Art. 236, V«X \b = \ab. REDUCTION OF RADICALS. TO REDUCE RADICALS OF DIFFERENT DEGREES TO EQUIVALENT RADICALS OF THE SAME DEGREE. 267. 1. Reduce y 1 2, ^3, and ^ 5 to equivalent radicals of the same degree. By Art. 252, ^2 = 2* = 2& =^2" = ^64 {f 3 = 3^ = 3^ = v^ 4 = y'Sl ^5 = 5* = 5& = y' 5 3 = v' 125 RADICALS. 189 RULE. Express the radicals 10 it h fractional exponents ; reduce these fractions to a common denominator • express the resulting fractional exponents with radical signs; and, finally, reduce the quantities under the radical signs to their simplest forms. Note. This method affords a means of comparison of the relative mag- nitudes of two or more radicals ; thus, in Example 1, as y/ 125 is evidently greater than y/Sl, and y'Sl than y/64, hence A'o is greater than y/3, and ^3 than y/2. EXAMPLES. Reduce the following to equivalent radicals of the same degree : 2. s/3, \f4, and ^5. 5. \f2~^ \J3~b, and ^4^ 94 ** 6, 4. 3. y 5, y 6, and y 7. 6. y a + b and y a — b. .3/ -, .*/ „ . /— 5 ; -■ 3/- 4. SJxy, y x z, and yyz. 7. y'cr — x 2 and sj a z — X s . 8. Which is the greater, ^3 or y'S? 9. Which is the greater, </2 or y/3? 10. Which is the greater, ^4 or $5 ? TO REDUCE RADICALS TO THEIR SIMPLEST FORMS. 268. A radical is in its simplest form when the quantity under the radical sign is not a perfect power of the degree denoted by any factor of the index of the radical, and has no factor which is a perfect power of the same degree as the radical. CASE I. 269. When the quantity under the radical sign is a perfect power of the degree denoted by some factor of the index of the radical. 190 ALGEBRA. a 1. Reduce y 8 to its simplest form. EXAMPLES. Reduce the following to their simplest forms : 2. #9. 4. ^27. 6. \/~a^W. 3. ^25 a 2 . 5. $125 a 3 1> 9 . 7. <l 25 "' CASE II. 270. When the quantity under the radical sign has a factor which is a perfect power of the same degree as the radical. 1. Reduce \J 32 to its simplest form. V 32 = V 16x2 = (Art. 266) ^16x^2 = 4^2, Am. 2. Reduce y/54 a 4 x to its simplest form. \/5±a i x = \/2Ta s x2ax = % 27~a~ 3 X ^2~a~x = 3 a \/2~a~x~, Ans. RULE. Resolve the quantity under the radical sign into two factors, one of which is the greatest perfect power of the same degree us the radical. Extract the required root of this factor, and prefix the result to the indicated root of the other. Note. In case the greatest perfect power in the numerical part of the quantity cannot be readily determined by inspection, it may always be ob- tained by resolving the numerical quantity into its prime factors. Let it be required, for example, to reduce ^ 1944 to its simplest form. 1944 = 2x2x2x3x3x3x3x3 = 2 8 x3 5 . Hence, ^1944 - V / 23lT35 = V2 2 x 3 4 x y/6 = 18 ^6. RADICALS. 191 EXAMPLES. Reduce the following to their simplest forms 3. ^50. 6. ^320. 9. 7^63aH 5 c 6 . 4. 3^24. 7. 2^80. 10. %250x 3 fz». 5. si 12. 8. \j98a s b 2 . 11. <Jl8x 3 y i -27 x* y 12. \/ax 2 — 6ax + 9a. 14. \/20 a x 1 + 60 a 2 x + 45 a 3 . 13. SJix^-y 2 ) (x + y). 15. ^192 a 4 6 5 + 320 a 3 6 4 . When the quantity under the radical sign is a fraction, mul- tiply both terms by such a quantity as will make the denomi- nator a perfect power of the same degree as the radical. Then proceed as before. /2 16. Reduce t / - to its simplest form. V / i=\/!=V / (H=Y / i x Y /6 =^ G '- te - /9 17. Reduce t / - to its simplest form. Reduce the following to their simplest forms : 18. 19 20 192 ALGEBRA. TO REDUCE A RATIONAL QUANTITY TO A RADICAL FORM. 271. 1. Eeduce 3 a; 2 to a radical of the third degree. 3 x 2 = \7 (3 x 2 ) 3 = \]21 x\ Ans. RULE. Raise the quantity to the power indicated by the given root, and write it under the corresponding radical sign. EXAMPLES. Reduce the following to radicals of the second degree : a rr a 3x x — 3 2. i a. 3. -=- . 4. a + 2 x. 5. . 5 x — 2 6. Reduce -=- to a radical of the fourth degree, o TO INTRODUCE THE COEFFICIENT OF A RADICAL UNDER THE RADICAL SIGN. 272. 1. Introduce the coefficient of 2 a y 3 x 2 under the radical sign. 2 a V 3a- 2 = \/ 8 a 3 X V 3 x 2 = (Art. 266) \f 8 a 3 X 3 x 2 = y 7 24 a 3 x% Ans. RULE. Reduce the coefficient to the form of a radical of the given degree; multiply together the quantities under the radical signs, and write the product under the given radical sign. EXAMPLES. Introduce the coefficients of the following under the radical signs : 2. 3^5- 4. 4a 2 \/^U. 6. 5c$Ja. 3. 2^7. 5. 3yTT^. 7. ( x -l)J(?±^). RADICALS. 193 ADDITION AND SUBTRACTION OF RADICALS. 273. 1. Find the sum of y/18, \J21, J -, and 12 y/^ 18 By Art. 270, . ^18 = 3^2 v/27= 3^3 12 2. Subtract ^48 from ^162. By Art. 270, ^162 = 3^6 ^48 =2^6 ^6, ^ws. RULE. Reduce each radical to its simplest form. Combine the similar radicals, and indicate the addition or subtraction of the dissimilar. EXAMPLES. Add together the following radicals : 5 3. y'S, ^18, and y/50. 6. ^20, t/i and J 4. ^12, v/48, and v/ 108. 7. y/|y/|> and y^ 5. ^16,^/54, and v' 128. 8. t/i'tf^**^ 2 27 7 2 3 194 ALGEBRA. Subtract the following : 9. v/ 45 from ^ 245. 10. J ~ horn J Simplify the following : 16 15 11. \/243 a b 2 + V 75 a* +^3a s - 54 a 2 b + 243 a b 2 . 12. 7^27-^75-y/| + v/12-y/l-y/l. 13. {^16 + 5^54-^250-^/^ + ^/1 + ^/^. . 15. V 63 « 2 « -S4:abx + 2Sb 2 x — ^7d 2 x + 42abx + 63b 2 x. MULTIPLICATION OF RADICALS. 274. 1. Multiply sj 2 by ^ 5. ^2x^5 = (Art. 266) ^2 X 5 = ^10, Arts. 2. Multiply y' 2 by y' 3. Reducing to equivalent radicals of the same degree (Art. 267), we have y/2 X $3 = $8 X \j 9 = ^72, Am. KULE. Reduce the radicals, if necessary, to equivalent ones of the same degree. Multiply together the quantities under the radi- cal signs, and write the product under the common radical sign. RADICALS. 195 EXAMPLES. Multiply together the following : 3. y/ 12 and y/ 3. 6. V 6 a* and \/ 5 a?. 4. ^2 and y^. 7. V^i V^ 4 , and V^^ 6 ) ' 5. ^axand^bx. 8. y/ 2, y/ 5, and t/ ^ • 9. Multiply 2y/cc — 3y/?/ by 4y/a: + y/?/. 2 y/cc — 3 \J y 4y/a;+ y/y 8ic — 12 \jxy + 2v/^7-3 2/ 8 « — 10 v/a; y — 3 ?/, ^4ras. Note. It should be remembered that to multiply a radical of the second degree by itself is simply to remove the radical sigu ; thus, y/ x x y/ x = x. Multiply together the following : 10. s/x- 2 and y]x + 3. 11. 3y/z - 5 and 7 sjx- 1. 12. s/.T + l-V/z-land y'a; + 1 + y/x-1 (Art. 106). 13. yV — 1 — a and V «' 2 — 1 + «• 14. \J x — s] y+ \J z and \j x -\- \J y — \/ z. 15. y/2-y/3 + y/5 and ^2 + ^3 + ^5. 16. 3y/5-2y/6 + y/7 and 6 y/5 + 4 y/6 + 2 y/7. 17. 4y/3-5y/2-2y/5 and 8 y/3 + 10 y/2 - 4 y/5. 196 ALGEBRA. Simplify the following : Square the following (Arts. 104 and 105) : 20. 2^3-^/2. 22. \/l-a 2 +a. 21. 3^8 + 5^3. 23. ^^b-\{a~+~b. DIVISION OF RADICALS. 275. Since (Art, 266), \faXs/b = \fab, it follows that \a b -i-tya = y / 6. RULE. Reduce the radicals, if necessary, to equivalent ones of the same degree. Divide the quantities under the radical signs, and write the quotient under the common radical sign. EXAMPLES. 8/ 1. Divide ^ 15 by y/ 5. Reducing to equivalent radicals of the same degree, we have ^15-^5 = ^225-^125 = ^/11 = ^/1, Ans. Divide the following : 2. v' 108 by v' 18. 6. s/2by$/3. 3. V^O^by \l~2Z. ■ 7, % 2 by ^3. 4. v/54b yv /6. 8. s/ 12 hjsj 2. 5. S/lT^by $3~o~. 9. \/Ta~by$Ta~. RADICALS. 197 INVOLUTION OF RADICALS. 276. 1. Raise fy 2 to the fourth power. (^ 2) 4 = (2*) 4 = 2* = f 2 4 = ^ 16, Jn». 2. Raise y/ 3 to the third power. (ft 3) 3 = (3") 3 = 3^ = 3* = \J 3, Ans. We observe that in the first example the quantity under the radical sign is raised to the required power ; while in the sec- ond, the index of the radical is divided by the exponent of the required power. Hence the following RULE. If possible, divide the index of the radical by the exponent of the required poioer. Otherwise, raise the quantity tinder the radical sign to the required poioer. Note. If the radical has a coefficient, it may be involved separately. The final result should be reduced to its simplest form. EXAMPLES. 3. Raise y/5 to the third power. 4. Square \J1. 5. Find the fourth power of 4 y 3 x. 6. Find the sixth power of y/ a 2 x. 7. Raise \l a — b to the fourth power. 8. Raise 3 a\bx to the fourth power. 9. Find the value of (\/ x + l)\ 10. Find the square of 4 V^ 2 — 3. 198 ALGEBRA. EVOLUTION OF RADICALS. 277. 1. Extract the square root of y 6 x 2 . \J( y^) = (y 7 ^ 2 )* = {(6 x*)*} * = (6 a; 2 )* = \/~6x~% Ans. 2. Extract the cube root of \/27 x 3 . V (V27^Q = (V27V)*= {v/(3 z) 3 P~ = {(3 a>)*}* = ( 3 a ')" = y 3 £, ^4?zs. We observe that in the first example the index of the radical is multiplied by the index of the required root ; while in the second, the required root is taken of the quantity under the radical sign. Hence the following RULE. If possible, extract the required root of the quantity under the radical sign. Otherwise, multiply the index of the radical by the index of the required root. Note. If the radical has a coefficient, which is not a perfect power of the same degree as the required root, it should he introduced under the radical sign hefore applying the rule. Thus, y(i\lax ) = y(\?l6ax ) = ylGax. The final result should be reduced to its simplest form. EXAMPLES. 3. Extract the square root of y/2. 4. Find the cube root of y/ 8. 4 5. Find the cube root of \a + b. 6. Find the square root of \x 2 — 2 x + 1. 7. Extract the fifth root of y/32. 8. Extract the cuhe root of y/27. 9. Find the value of ^(3 y/3). RADICALS. 199 5/ 10. Find the fourth root of \ x* y vi . 11. Find the value of \J(±\J2). TO REDUCE A FRACTION HAVING AN IRRATIONAL DENOMINATOR TO AN EQUIVALENT ONE WHOSE DENOMINATOR IS RATIONAL. CASE I. 278. When the denominator is a monomial. 2 b 1. Reduce -; — to an equivalent fraction whose denominator sj a is rational. Multiplying both terms by y/ a, 2b _2bsja _2bsja \l a y/ a\j a a , Ans. 5 2. Reduce ^-^ to an equivalent fraction whose denominator is rational. Multiplying both terms by y/ 9, 5_ 5^9 _5v^9_5^9 RULE. Multiply both terms of the fraction by a radical of the same degree as the denominator, with such a quantity under the radical sign as will make the denominator of the resulting fraction rational. 200 ALGEBEA. EXAMPLES. Reduce the following to equivalent fractions with rational denominators : 3 3 3< ^2" 4 1 $2 a 5 5 5 ' IN- CASE II. 6- r 2c V '27 a 2 279. When the denominator is a binomial, containing only radicals of the second degree. 1. Reduce — — - r^ to an equivalent fraction whose denomi- nator is rational. Multiplying both terms by 3 — y/ 2, 10 10( 3-^2) fK ^ iA ^ 30-10y/2 3T72 = (3 + V2)(3-^2) = ^ Art 106 > 7 ' ^ 2. Reduce — —- to an equivalent fraction whose de- nominator is rational. Multiplying both terms by ^ 5 + \J 2, v/5 + v /2 _ ( v /5 + v/2)(y/5 + v/2) _ 7 + 2 v /10 v/5-v/2 _ ( v /5- s /2)( v /5 + v /2)~" 3 ; AnS ' RULE. Multiply both terms of the fraction by the denominator with the sign between its terms changed. EXAMPLES. Reduce the following to equivalent fractions with rational denominators : RADICALS. 201 4 2y/5 + y/2 s/a + x+^a-x 3< 3TV2* " s/5-3^2' • y/^r^-y^T a; , 4-y/3 y/«-y^ 19 yV-l-y/^+1 2-v/3 s/a+s/x ^at-l+^at+l _ VS-V 73 o 2 +V« + 1 1Q ^ + V/^ 2 -4 «• ~777 To • "• , • A«5- , • V/2 + v/3 1-Va+l » — VaJ 2 -4 y/ft + y/5 1Q « — V/V— 1 14 y/x — 4 y/a — 2 sja-\jb' ' a + yjd 1 —! ' 2\lx + 3\jx — 2* 280. If the denominator is a trinomial, containing only radicals of the second degree, by multiplying both terms of the fraction by the denominator with one of its signs changed, we shall obtain a fraction which can be reduced to an equiva- lent fraction with a rational denominator by the method of Case II. Thus, to reduce the fraction y/2-v/3-v/7 ^2 + ^3 + ^' Multiplying both terms byy/2 + y/3 — y/7, v /2- v /3-v/7 _ (v/2-v/3-v/7)(v/2 + v /3-v/T) _6-2 v 14 l/2 + y/3 + y/7~(\/2 + s/3 + \/7)(s/2+sJ3-s/7) 2y/6-2 _ 3-y/14 Multiplying both terms by y/ 6 + 1, we Lave (3- V /14)( v /6 + l) _ 3-v/14 + 3y/6- v /84 (v / 6 _l )(v /6 + l) - 5 If the denominator is a binomial, containing radicals of any degrees whatever, it is possible to reduce the fraction to an equivalent form with a rational denominator ; but the process is more complicated than the preceding and rarely necessary. 202 ALGEBEA. 281. To find the approximate value of a fraction whose denominator is irrational, reduce it to an equivalent fraction whose denominator is rational. 1. Find the value of ^ j^ to three decimal places. 2 -J^ = (Art. 279) 2 -±^- 2 = 2 -±|^ = 1.707, An*. It will be seen that the value of the fraction is obtained in this way more easily than by dividing 1 by 2 — \J 2, or its value .586. EXAMPLES. Find the values to three decimal places of the following : 2 2 3 3 4 - 5 V /3 -V /2 IMAGINARY QUANTITIES. 282. An Imaginary Quantity is an indicated even root of a negative quantity ; as, y — 4, \/ — a 2 . In contradistinction, all other quantities, rational or irra- tional, are called real quantities. 283. All imaginary quantities may be expressed in one common form, which is, a real quantity multiplied by y — 1. For example, y/3^ _ yV X (_i) _ (Art. 266) y' a" X ^ :Z 1 = a y^l ; also, y/- 2 = ^2 X(-l) = s/2\[^l. Hence, we may regard ^—1 as a universal factor of every imaginary quantity, and use it in our investigations as the only symbol of such a quantity. KADICALS. 203 284. Imaginary quantities may be added, subtracted, and divided the same as other radicals ; but with regard to multi- plication, the usual rule requires some modification. 285. By Art. 17, V— 1 means such an expression as when multiplied by itself produces — 1 ; or,- (v/^l7 = -l; also, (y/I^^v/^lTxV^T^-lV 3 !; and, (V^1) 4 = (V^1) 2 X(V^ 3 1) 2 =(-1)X(-1) = 1. By continuing the multiplication, we should find (V-l) 8 = l; etc. Or, in general, where n is any positive integer, (v/=ir + W=T; (V=i) 4B+2 =-i; (V=T) 4n+8 =-V^; (V=1) 4 " +4 = L MULTIPLICATION OF IMAGINARY QUANTITIES. 286. 1. Multiply ^^a^hy \f^¥. V^^X V 17 ^^ (Art. 283) a <f^l X ft y'^1 = « ft (V 11 *) 2 = — aft, ^4hs. 2. Multiply V- 2 by y^3. V^2 X y /r 3=v/2x\/3x(v /:: i) 2 =-^ ) il ??s. 3. Multiply together ^— 4, V— 9, V~ 16 > and V" 25. V^ X V 3 ^ X S/^IG x V /Ir 25 = 2x3x4x5x (v^)* = 120(\/~l) 4 = 120, Arts. 204 ALGEBRA. RULE. Reduce all the imaginary quantities to the form of a real quantity multiplied by ^— 1. Multiply toy ether the real quantities, and multiply the result by the required power of EXAMPLES. Multiply the following : 4. 4^-3 and 2\J-2. 7. 1 + V- 1 and 1 - \J- 1. 5. _ 3 yL_ a and 4 V- b. 8. \J- a% V~ b% and y— c 2 . 6. 4 + V- 7 and 8-2^-7. 9. a + \/— b and a — \J—b. 10. 2 V- 3 — 3 V- 2 and 4 V- 3 + 6 V~ 2. 11. Divide V- « by V~ *• We should obtain the same result by using the rule of Art. 275 ; hence, that rule applies to the division of all radicals, whether real or imaginary. Divide the following : 12. V- 6 by V- 2. 13. ^^12 by y'-S. Simplify the following : 16. 1 + S ^_1 . (Art. 279). 1-V-l 2-V-2 14. V- -5hj^- 1. 15. tf- - 54 by y 7 - 17. A + t -2 18. ^£0 + "-^ . (Art. 154). a — \'—b a +^-b 19. Expand (2 -V^3) a . 20. Expand (2 + 3 Y^2) 8 . KADICALS. 205 QUADRATIC SURDS. 287. A Quadratic Surd is the indicated square root of an imperfect square ; as, \J 3, ^ x + 1. 288. A Binomial Surd is a binomial in which one or both of the terms are irrational. 289. The square root of a rational quantity cannot be equal to a rational quantity plus a quadratic surd. If possible, let y' a = b + y/ c Squaring the equation, a = b 2 + 2 b \j c + c or, 2 b \J c = a — b 2 — c . a — b 2 — c ^ C = ^b— that is, a surd equal to a rational quantity, which is impossible. 290. If tin' sum of a rational quantity and a quadratic surd be equal to the sum of another rational quantity and another quadratic surd, the two rational quantities will be equal, also the two quadratic surds. That is, if a + \J b — c + \J d then a = c and \Jb = \J d For, if a is not equal to c, suppose a = c -f- x then c-\-x-\-\Jb = c + \Jd or, x + s/ b = y/ d which is impossible by Art. 289. Hence, a must equal c, and consequently y/ b must equal y/ d. 291. To prove that ifsja + ^b — sj x + \J y, then V ' a — \j b = s /x — sfy. Squaring the equation \/ a + sjb = \J x + \J y, we have tc + \Jb — x + 2 \l~r~y + y Whence, by Art. 290, a = x + y and >Jb = 2 \ xy. 206 ALGEBRA. Subtracting these two equations, we have a~sJb — x — 2sJxy-\-y Extracting the square root, ^ a — ^b = ^ x — ^ y. 292. To extract the square root of a binomial surd whose first term is rational. For example, to extract the square root of a + \J b. Assume \J a + \fb = \/ x + sj y (1) then (Art. 291), ^a—sjb = ^ x — \Jy (2) Multiplying (1) by (2), \ja"-b = x-y (3) Squaring (1), a + \/b — x + 2^xy + y Whence (Art. 290), a = x + y. (4) Adding (3) and (4), a + \J~aT^b = 2x, or x= l + ^f~ b - Subtracting (3) from (4), a — SJ a 2 — b=2y, or ?/= ^- . Substituting these values of x and y in (1) and (2), ^i^=^(l±^EI) + v /(»=^E»). (5) EXAMPLES. 1. Find the square root of 3 + 2 ^ 2 or 3 + y/ 8. Here a = 3 and b = 8. Substituting in (5), we have V 3T^=v/( 3 -±^) + V /('^^) = v /(^) +v /^H 2 + i >-<- RADICALS. 207 2. Find the square root of 6 — \J 20. Here a = 6 and b = 20. Substituting in (6), we have 293. Examples of this kind may always he solved hy the following method : 3. Extract the square root of 14 — 4 y/ 6. y , 14_4 v /6 = V / l^-2v /24 = V / 12-2v / 24 + 2 = (Art. 116)^12-^2 = 2^3-^2, Arts. 4. Extract the square root of 43 + 15 \J 8. V43 + 15 si 8 = \M3 + sj 1800 = ^43 + 2 sj 450 = ^25 + 2^450 + 18 = v /25 + V /18 = :5 + 3 v/ 2 > ^ s - EULE. Reduce the surd term so that its coefficient may he 2. Sep- arate the rational term into tiro parts whose product shall be the quantity under the radical sign (see first note on page 48), writing one part he/ore the surd term and the other part after it. Extract the square roots of these parts, and connect them by the sign of the surd term. The advantage of this method is that it does not require the memorizing of formulae (5) and (6). EXAMPLES. Extract the square roots of the following : 5. 12 + 2^35. 8. 35 + 10 <J 10. 11. 20-5^12. 6. 24-2^63. 9. 12 -sj 108. 12. 14 + 3^20. 7. 16 + 6^7. 10. 8-v/60. 13. 67 -7 si 12. 208 ALGEBRA. Extract the square roots of the following, using formula? (5) and (6), Art. 292: 14. l-12\/-2. 15. 7 + 30V-2. 16. 35-3V/-16. 17. 2m-2\Jm--n\ 18. x 2 + a x -2 \/ax 3 . Extract the fourth roots of the following : 19. 193 + 22 y/ 72. 20. 17-12)/ 2. 21. 97-56^3. SOLUTION OF EQUATIONS CONTAINING RADICALS. CASE I. 294. Wlien there is only one radical term in the equation. 1. Solve the equation v/ar 2 — 5 — x = — 1. Transposing, ^ x 2 — 5 = x — 1 Squaring, x 2 — 5 = x 2 — 2 cc + 1 Whence, x = 3, Ans. CASE II. 295. When there are two radical terms in the equation. 2. Solve the equation \] x — \ f x — 3=1. Transposing, \J x — 1 = ^ x — 3 Squaring, x — 2\Jx + l = x — 3 Transposing and uniting, — 2 y/ x = — 4 or, \jx = 2 Whence, x = 4, Ans. CASE III. 296. When there are three radical terms in the equation. 3. Solve the equation \J x + 6 + \J x + 13 — v/4a; + 37 = 0. RADICALS. 209 Transposing, \J x + 6 + \/x + l§ = \/4:X + 3T Squaring, x + 6 + 2 \/cc 2 + 19.r + T8 + x + 13 = 4 x + 37 Transposing and uniting, 2 y as 2 + 19 a; + 78 = 2 a; + 18 or, ^+19 a; + 78 = a: + 9 Squaring, a; 2 + 19 x + 78 = a;' 2 + 18 a; + 81 Whence, x = 3, ^f «s. RULE. 297. Transpose the terms of the given equation so that a radical term may stand alone in one member ; then raise each member to a power of the same degree as the radical. If there is still a radical term remaining, repeat the op- eration. The equation should he simplified as much as possible hefore performing the involution. Note. All the examples in tliis chapter reduce to simple equations ; radical equations, however, may reduce to equations of the second degree, for the solution of which see Chapter XXIV. EXAMPLES. Solve the following equations : 3/ 4. V«-8 = 3. 6. y'3a; + 4 + 3 = 6. 8. 8-2^ x 5. V^-3 = 2. 7. ^3-1-2 = 1. 9. o~\2x = 3. 10. y/4ar-19-2a- = -l. 14. 6 + ^x = \Jl2 + 11. ^^-3^+6-1=1-^. 15. V / -'-32 + v /a-r=16. 12. yx* — 6x 2 + 2 = x. 16. ^x — 3— Va; + 12=— 3. 13. ^ + ^ + 5 = 5. 17. \l2x-l+\J2x + 9 = 8. 18. ^3x + 10-^3x + 25 = -3. 19. ^x 2 -3x + 5-\'x 2 -5x-2 = l. 210 ALGEBEA. 20. \Jx 2 + 4 x + 12 + \jx 2 - 12 x - 20 = 8. 21. v^— ^-3 = — . y x 22. ^3^+^3^+13 = ^ ==. V o x + 13 23 V 7 ^ — 3 _y/g; — 4 V /tc + T _ "y'ic + 1' 24 y/^_+38_ y/a; + 28 \/# + 6 \J x + 4 ' 25. ^-1 + ^ + 4=^4^ + 5. 26. yWl + V*-2-V / 4jc-3 = 0. 27. \j2x-3- ^8^ + 1 + VlS ^-92 = 0. 28. y^ - 3 - V^ - 14 - y/4 a - 155 = 0. 29. x- \^ (9 + x \/^~~3) = 3. 30. x + l = \f(l + x \/^ r +T6). 31 v5^y/3_v^+_3 V / 2^"- v /2~V / ^ + 2' 32. y( a *-3a>x + x 2 \/3^~x) = a-x. XXIV. — QUADRATIC EQUATIONS. 298. A Quadratic Equation, or an equation of the second degree (Art. 164), is one in which the square is the highest power of the unknown quantity ; as, ax 2 = b, and x* + 8 x = 20. 299. A Pure Quadratic Equation is one which contains only the square of the unknown quantity ; as, ax- = b; and a; 2 = 400. QUADEATIC EQUATIONS. 211 Equations of this kind are sometimes called incomplete equa- tions of the second degree. 300. An Affected Quadratic Equation is one which con- tains both the square and first power of the unknown quan- tity ; as, x 2 + 8 x = 20 ; and a x 2 + b x — e = b x 2 — a x + d. Equations of this kind, containing every power of the un- known quantity from the first to the highest given, are some- times called complete equations. PURE QUADRATIC EQUATIONS. 301. A pure quadratic equation can always he reduced to the form x 2 = a, in which a may represent any quantity, positive or negative, integral or fractional. Thus, in the equation 20a; 2 , K 9 Jv 41 3-5x 2 Clearing of fractions, 80 x 2 - 12 (5 x 2 + 4) = 41 - (9 - 15 x 2 ) or, 80 x 2 - 60 x 2 - 48 = 41 - 9 + 15 x 2 Transposing and uniting terms, 5 x 2 = 80 x 2 = 16 which is in the form x 2 = a. Equations of this kind have, therefore, sometimes been de- nominated binomial, or those of two terms. 302. An equation of the form x 2 = a may he readily solved by taking the square root of each mem- ber. Thus, x = ± \Ja, 212 ALGEBKA. where the double sign is used, because the square root of a quantity may be either positive or negative (Art. 237). Note. It may seem at first as though we ought to write the douhle sign before the square root of each member, as follows : ±x = ± y/a. We do not omit the double sign before the square root of the first member because it is incorrect, but because we obtain no new results by consid- ering it. The equation ± x = ± y/ a can be written in four different ways, thus, x = ^a — x=tfa -x— - y/« where the last two forms are equivalent to the first two, and become iden- tical with them on changing all the signs. Hence it is sufficient, in extracting the square root of both members of an equation, to place the double sign before one member only. 5x 2 303. 1. Solve the equation 3 x 2 + 7 = —r- + 35. Clearing of fractions, 12 x 2 + 28 = 5 x 2 + 140 Transposing and uniting terms, 7 x 2 = 112 x 2 = 16 Extracting the square root of both members, x = ± 4, Ans. RULE. 'Reduce the given equation to the form x 2 = a,and then extract the square root of both members. EXAMPLES Solve the following equations : 2. 4a; 2 - 7 = 29. 4. 3. 5x 2 + 5 = 3x 2 +5o. 5. t X 1 — ■5 = Sx 2 -11 5 8 "3 5 4 + x 4 — X QUADRATIC EQUATIONS. 213 245 x = 5x. 7. 13-V/3a; 2 +lG = 6. G 8. x +^x* + 3= ,-z—z y x- + 6 9. _y/3 l_y/l_aj a 1 + y/l- a- .'• _ cc 2 5 a; 2 _ 7 2 335 10, Y~ + U^2l~ X + ~2l' 11. 2 (x - 3) (x + 3) = (x + l) 2 - 2 x. 12. aa; 2 + 5 = c. 13 x 2 — b x 2 — a AFFECTED QUADRATIC EQUATIONS. 304. An affected quadratic equation may always be reduced to the form x 2 -\-jpx = q, where p and q represent any quantities, positive or negative, integral or fractional. Thus, in the equation 3x — 3 _ 3x — 6 5 x 5- = 2 x H s — x — J Clearing of fractions, 10 x (x _ 3) _ (6 x - 6) = 4 x (x - 3) + (3 x - 6) (a; - 3) or, 10 a- 2 - 30 x - 6 x + 6 = 4 x 2 — 12 a: + 3 a 2 - 15 a; + 18 Transposing and uniting terms, 3 a; 2 — 9 x = 12 Dividing by 3, x 2 — 3 x = 4 which is in the form x 2 + p x =■ q. Equations of this kind have, therefore, sometimes been de- nominated trinomial, or those of three terms. 214 ALGEBRA. 305. Let it be required to solve the equation x 2 -\-px = q. Equations of this kind are solved by adding to both members such a quantity as will make the first member a perfect square, and taking the square root of the resulting equation. The pro- cess of adding such a quantity to both sides as will make the first member a perfect square, is termed Completing the Square. In any trinomial square (Arts. 104 and 105), the middle term is twice the product of the square roots of the extreme terms ; therefore the square root of the last term must be equal to half the second term divided hy the square root of the first. Hence the square root of the quantity which must be added to x~ + p x to render it a perfect square, is ~- —- x, or — . Adding to both members the square of ^, or ^-, we have 2 , _,, , P , p _±q+p X<+px + - = q + ^ = - Extracting the square root of both members, X + P = ± \/±<I+P' or, x = — 7 -± p ^iq+jr 2 - 1 - 2 Thus, there are two values of x, p \ 4: q + p' 2 ■ p SJ 4 q + p 2 x = -t>+ —o i or 2 ' 2 ' 2 2 We observe from the preceding investigation that the quan- tity to be added to complete the square is found by taking half the coefficient of cc, and squaring the result. Hence, for solving affected quadratic equations, we have the following QUADKATIC EQUATIONS. 215 RULE. Reduce the equation to the form x 2 + p x = q. Complete the square by adding to both members the square of half the coefficient of x. Extract the square root of both members, and solve the simple equation thus found. 1. Solve the equation x 2 — 3 x = 4. Completing the square, by adding to both members the .3 9 square of -, or -, o 9 . 9 25 x 2 — 3x + - = 4:+ - = — 4 4 4 ■^ 3 5 Extracting the square root, x — ~ = ± ^ ™ 3 5 Transposing, a; = - ± - Taking the upper sign, x = -r+ k = h — 4. Z ju Li 3 5 2 Taking the lower sign, x= 7> — ^ = — 7 , = — 1. Ll Li Li Ans. x = £ or — 1. We may verify these values as follows : Putting x = 4 in the given equation, 16 — 12 = 4. Putting x = -l, 1 + 3 = 4. These results being identical, the values of x are verified. 2. Solve the equation 3 x 2 + 8 x = — 4. 8x4 Dividing through by 3 x 2 + -— -=— - o o 216 ALGEBRA. Completing the square, by adding to both members the square ,4 16 0f 3' 0r y 2 8^ 16 _ 4 16_4 4 2 Extracting the square root, x + - = ± - o o m 4 2 1 ransposing, x = — - ± - rr i • n 4 2 lakmg the upper sign, x = — - + - = — - Taking the lower sign, x = — ^ — ^ = — ^ = — 2. 2 .4ns. cc = — - or — 2. o 3. Solve the equation — 3x 2 — 7 x = -=-. o 7 x 10 Dividing through by — 3, x 1 + -~- = — — Completing the square, by adding to both members the square . 7 49 of 6'° r 36' X 2 + 7x 49 10 49 __ 9 X + 36" ~~9~ + 36~36 Extracting the square root, 7_ 3 X ~t~ „ it A b b Transposing, 7 3 X ~ 6 ± 6 Whence, 2 5 A x = — q or — k ) -4«& «5 O A SECOND METHOD OF COMPLETING THE SQUARE. 306. Although any affected quadratic equation may be solved by the method of Art. 305, since its rule is general, QUADRATIC EQUATIONS. 217 still it is sometimes rilore convenient to employ a second method of completing the square, known as the " Hindoo Method." An affected quadratic, reduced to three terms, and cleared of all fractions, may he reduced to the form a x 2 + b x = c. Multiplying each term by 4 a, we have By an operation similar to that of Art. 305, we may show that b 2 must be added to both members, in order that the first member may be a perfect square. Thus, 4 a 2 x- + 4 a b x + b 2 = b 2 + 4 a c Extracting the square root, 2 ax -\- b = ± \ b 2 -\- 4ta c Transposing, 2 ax = — b ± \ b 2 + 4 a c tv -a- i q -h±\?b 2 +4ac Dividing by 2 a, x = ^ . It will be observed that the quantity necessary to complete the square, is the square of the coefficient of x in the given equation. Hence the following RULE. Reduce the equation to the form a x 2 + b x = r. Multiply both members of the equation by four times the coefficient of x 2 , and add to each the square of the coefficient of x in the given equation. Extract the square root of both members, and solve the sim- ple equation thus produced. Note. The only advantage of this method over the preceding is in avoiding fractions in completing the square. 4. Solve the equation 2 x 2 — 7 x = — S. 218 ALGEBRA. Multiplying both members by four times 2, or 8, 16 x 2 -5Gx = - 24 Adding to each member the square of 7, or 49, 16 x 2 — 56 x + 49 = - 24 + 49 = 25 Extracting the square root, 4 x — 7 = ± 5 Transposing, 4cc = 7±5 = 12or2 Dividing by 4, x = 3 or - , ^4«s. 307. This method is usually to be preferred in solving literal equations. 5. Solve the equation x 2 + (a — 1) x = a. Multiplying both members by four times 1, or 4, 4 x 2 + 4 (a — 1) x = 4 a Adding to each member the square of a — 1, or (a — l) 2 , 4 x 2 + 4 (a - 1) x + (a - l) 2 = 4 a + (a - l) 2 = a 2 + 2 a + 1 = (a + l) 2 Extracting the square root, 2 a; + (a — 1) = ± (a + 1) Transposing, 2 cc = — (a — 1) ± (a + 1) Taking the upper sign, 2 x = — (a — 1) + (a + 1) r= — « + l + «+l=2 or, x = l. Taking the lower sign, 2 x = -(a-l)-(a + l) = — a + 1 — a — 1 = — 2 a or, x — — a. Ans. x = 1 or — a. 308. In case the coefficient of x in the given equation is an even number, the rule may be modified as follows : QUADRATIC EQUATIONS. 219 Multiply both members of the equation by the coefficient of x 2 , and add to each the square of half the coefficient of x in the given equation. 6. Solve the equation 7 x 2 + 4 x = 51. Multiplying both members by 7, 49 x 2 + 28 x = 357 Adding to each member the square of 2, or 4, 49 x 2 + 28 x + 4 = 361 Extracting the square root, 7 x + 2 = ± 19 Transposing, 7 a; — — 2 ± 19 = 17 or — 21 17 Dividing by 7, ic = — or — 3, ^l»s. SOLUTION OF QUADRATIC EQUATIONS BY A FORMULA. 309. In Art. 30G, we showed that if a x 2 + b x = c, then -b± \Jb 2 + ±ac ... X = 2a * (1) We may use this as a formula for the solution of quadratic equations as follows : 7. Solve the equation 3 x 2 + 5 x = 42. Here a = 3, b = 5, c = 42 ; substituting these values in (1), -5±\/25 + 504 X — 6 -5±v/529_-5±23 8. Solve the equation 110 « 2 — 21 x = — 1. Here a = 110, 6 = — 21, c = — 1 j substituting in (1), 21 ±1^441 -440 21 ±1 1 1 . X = ~ -22CT - = -220- = I6 Or ll'^ 220 ALGEBRA. Note. Particular attention must be paid to the signs of the coefficients in substituting. 9. Solve the equation, — x 2 — 6 x = 8. Here a = — 1, b = — 6, c = 8; substituting in (1), 6 ±^36- 32 6±2 x = —jr = — = — 4 or — 2, Ans. RULE. Reduce the equation to the form a x 2 + b x = c. The value of x is then a fraction, ivhose numerator is the coefficient of x with its sign changed, plus or minus the square root of the sum of the square of said coefficient, and four times the product of the second member by the coefficient of x' 2 ; and whose denominator is twice the coefficient of x 2 . 310. The following equations may he solved hy either of the preceding methods, preference being given to the one best adapted to the example considered. Special methods and devices may also be employed whenever any advantage can thereby be gained. EXAMPLES. Solve the following equations : 10. x°- + 2x + 7 = 4:2. 16. 26x + lox 2 = -7. 11. a; 2 - 9 x- 22 = 0. 17. - 40 + x = 6 x\ 12. x 2 -Sx = -15. 18. 17x = 2x 2 -6. 13. z 2 +18* = -65. 19. ^+*=_1. t 7 9 f 2 14. Gx*+7 X -3 = Q. 20. x =--^-. £ o o 3x 2 °2 15. 13 z - 14 = 3 a;'. 21. ~-^ = x . 5 o QUADRATIC EQUATIONS. 221 22. ^_-"-_£ = a 24. (as-3)(2as + l)=4. 6 Jo 23. ^--^ = -^. 25. (*+5)O-5)-(llx+l)=0. 26. 4x(18x-l) = (10x-l) 2 . 27. (3a:-5) 2 -(> + 2) 2 = -5. 28. (as-l) 2 -(3as + 8) 2 =(2as + 5) 2 . 29 2 as_ 5 _21 ^_ 3 4 29< x + 2-~2- d7 " 5-as 7 7* x x — 1_3 _ ft cc + 1 a? + 3 _8 30 - ^TI— "F""2' ^+2 x + 4 _ 3- .t 5 — as 15 on 3 a? 2 1 — 8 x x 31. -3 = -j- . OV. 5 — x . x 4' ' x — 7 10 5 5 3z + l 1 . n 2as-l 3x 1 . = - . 40. 1 — x a; 2 4" x 3 a; — 12 cc 33. = -— — -. 41. \/20 + x-x 2 = 2(x-5). 3x+4 4x+l T 34. _ x __— — - = 0. 42. as+V5as + 10 = 8. 3a; + 4 7 as- 4 ._ . 35-3as ,. AQ as*-a* + 7 n 35. 6 as H = 44. 43. = a; + — a; ar + 3 a; — 1 o 14 — r 7 3 22 36. 4 a -=t-,? = 14 44. -^ ^=^- a; + 1 x 2 — 4 a; + 2 o 45. ^— +- = — - - + x* — 1 3 3 (as — 1) a; + 1 222 ALGEBRA. ._ a + 3 x — 3 2 a; — 3 46. -s+- -s = r - a + 2 x — 2 x — 1 ,„ 35 + 2 a — 2 2a + 16 47. T H =-=- -=-, a — 1 a + 1 a + 5 12 + 5 a 2 + a 1 12 — 5a a 1 — 5a 49. \J X + 1 — ya- -1 a \Jx X ■+ 4/9 + 1 ■ V2 + v^- -1 "2* 50. (5x4- - V^ 3) = = 9. K1 a 2 + 36 a a 52. a ex 2 — b ex + a d x — b d. 53. a 2 - 2 a a + a 2 - & 2 = 0. 2 a (a — a) a 54 3a-2a 4 1 111 55. - = - + - + -. a + b + x a b a 56. (3 a - 2) (a + 5) - (a - 6) (5 a - 16) =301. 57. (2 a + 3) (3 a + 4) = (8 + a) (2 a + 9). 58. (2 a - 5) 2 - (2 a - 1) 2 = 8 a - 5 a 2 - 5. 59. x- + b x + c x = (a + c) (a — b). Sa-x 6 a 2 + a b - 2 b 2 b 2 x 60. abx 2 + c 2 61. (3 a 2 + b 2 ) (a 2 - a + 1) = (3 6 2 + a 2 ) (a 2 + a + 1). PROBLEMS. 223 XXV. — PROBLEMS LEADING TO PURE OR AFFECTED QUADRATIC EQUATIONS CONTAINING BUT ONE UNKNOWN QUANTITY. 311. 1. I bought a lot of flour for $ 175 ; and the number of dollars per barrel was to the number of barrels, as 4 to 7. How many barrels were purchased, and what was the price of each ? Let x = the number of dollars per barrel, 7 x then - = the number of barrels. 4 7 x 2 By the conditions, —r- = 175 Whence, x = ± 10. Only the positive value is applicable, as the negative value does not answer to the conditions of the problem. That is, x = 10, the number of dollars per barrel, 7 x and - = 171, the number of barrels. 4 2. There is a certain number, whose square increased by 30, is equal to 11 times the number itself. Required the number. Let x — the number. By the conditions, x 2 + 30 = 11 x Solving this equation, x = 5 or 6. That is, the number is either 5 or G, for each of these values satisfies the conditions of the problem. 3. I bought a watch, winch I sold for $ 56, and thereby gained as much per cent as the watch cost me. Required the amount paid for it. Let x = the amount paid, in dollars. Then x = the gain per cent, x X" and — — X x = — — - = the whole gain in dollars. 224 ALGEBRA. x 2 By the conditions, -^-r— = 56 — x J 100 Solving this equation, x = 40 or — 140. Only the positive value of x is here admissible, as the nega- tive result does not answer to the conditions of the problem. The cost, therefore, was $ 40. Note. When two answers are found to a problem, they should be ex- amined to see whether they answer to the conditions of the problem or not. Only those which answer to the conditions should be retained. PROBLEMS. 4. I have three square house-lots, of equal size. If I were to add 193 square rods to their contents, they would be equiv- alent to a square lot whose sides would each measure 25 rods. Required the length of each side of the three lots. 5. There are two square fields, the larger of which contains 25,600 square rods more than the other, and the ratio of their sides is as 5 to 3. Required the contents of each. 6. Find two numbers whose sum shall be 15, and the sum of their squares 117. 7. A person cut and piled two ranges of wood, whose united contents were 26 cords, for 356 dimes ; and the labor on each of them cost as many dimes per cord as there were cords in its range. Required the number of cords in each range. 8. A grazier bought a certain number of oxen for $ 240, and having lost 3, he sold the remainder at $8 a head more than they cost him, and gained $59. How many did he buy ? 9. The plate of a rectangular looking-glass is 18 inches by 12, and is to be framed with a frame all parts of which are of equal width, and whose area is to be equal to that of the glass. Required the width of the frame. 10. A merchant sold a quantity of flour for §.'59, and gained as much per cent as the flour cost him. What was the cost of the flour ? PROBLEMS. 225 11. There are two numbers whose difference is 9, and whose sum multiplied by the greater is 266. What are the numbers ? 12. A and B gained by trade $ 1800. A's money was in the firm 12 months, and he received for his principal and gain $2600. B's money, which was $3000, was in the firm 16 months. What money did A put into the firm ? 13. A merchant bought a quantity of flour for $ 72, and found that if he had bought 6 barrels more for the same money, he would have paid $> 1 less for each barrel. How many barrel's did he buy, and what was the price of each ? 14. A square courtyard has a gravel-walk around it. The side of the court wants 2 yards of being 6 times the breadth of the gravel-walk, and the number of square yards in the walk exceeds the number of yards in the perimeter of the court by 164. Required the area of the court. 15. My gross income is $ 1000. After deducting a percent- age for income tax, and then a percentage, less by one than that of the income tax, from the remainder, the income is reduced to 8 912. Required the rate per cent at which the income tax is charged. 16. The sum of the squares of two consecutive numbers is 113. What are the numbers ? 17. Find three consecutive numbers such that twice the product of the first and third is equal to the square of the second, increased by 62. 18. I have a rectangular field of corn which consists of 6250 hills ; and the number of hills in the length exceeds the num- ber in the breadth by 75. How many hills are there in the length and breadth ? 19. A certain company agreed to build a vessel for $ 6300 ; but, two of their number having died, those that survived had each to advance $ 200 more than they otherwise would have done. Of how many persons did the company at first consist ? 226 ALGEBRA. 20. A detachment from an army was marching in regular column, with C men more in depth than in front; hut Avhen the enemy came in sight, the front was increased by 870 men, and the whole was thus drawn up in 4 lines. Required the number of men. 21. A has two square gardens, and the side of the one ex- ceeds that of the other by 4 rods, while the contents of both are 208 square rods. How many square rods does the larger garden contain more than the smaller ? 22. A certain farm is a rectangle, whose length is twice its breadth; but should it be enlarged 20 rods in length and 24 rods in breadth, its contents would be doubled. Of how many acres does the farm consist ? 23. A square courtyard has a rectangular gravel-walk around it. The side of the court wants one yard of being six times the breadth of the gravel-walk, and the number of square yards in the walk exceeds the number of yards in the perim- eter of the court by 340. What is the area of the court and width of the walk ? 24. A merchant bought 54 bushels of wheat, and a certain quantity of barley. For the former he gave half as many dimes per bushel as there were bushels of barley, and for the latter 4 dimes per bushel less. He sold the mixture at $ 1 per bushel, and lost $ 57.60 by his bargain. Required the quan- tity of barley, and its price per bushel. 25. A lady wishes to purchase a carpet for each of her square parlors ; the side of one of them is 1 yard longer than the other, and it will require 85 sqxiare yards for both rooms. What will it cost the lady to carpet each of the rooms with carpeting 40 inches wide, at 81.75 per yard ? 26. A man has two square lots of unequal dimensions, con- taining together 15,025 square feet. If the lots were contigu- ous to each other, it would require 530 feet of fence to embrace them in a single enclosure of six sides. Required the area of each lot. QUADRATIC EQUATIONS. 907 27. A certain number consists of two digits, the left-hand digit being twice the right-hand ; and if the digits are inverted, the product of the number thus formed, increased by 11, and the original number, is 4956. Find the number. 28. A man travelled 108 miles. If he had gone 3 miles more an hour, he would have performed the journey in 6 hours less time. How many miles an hour did he go ? 29. A cistern can be filled by two pipes running together in 2 hours 55 minutes. The larger pipe by itself will fill it sooner than the smaller by 2 hours. What time will each pipe separately take to fill it ? 30. A set out from C towards D, and travelled 3 miles an hour. After he had gone 28 miles, B set out from D towards C, and went every hour ^ of the entire distance ; and after he had travelled as many hours as he went miles in an hour, he met A. Required the distance from C to D. 31. A courier proceeds from P to Q in 14 hours ; a second courier starts at the same time from a place 10 miles behind P, and arrives at Q at the same time as the first courier. The second courier finds that he takes half an hour less than the first to accomplish 20 miles. Find the distance from P to Q. XXVI. — EQUATIONS IN THE QUADRATIC FORM. 312. An equation is in the quadratic form when it is ex- pressed in three terms, two of which contain the unknown quantity ; and of these two, one has an exponent twice as great as the other. As, x 6 — 6 x 3 = 16, x s + x* = 72, (x- - 1) 2 + 3 {x- - 1) = 18, etc. 228 ALGEBRA. 313. The rules already given for the solution of quadratics will apply to equations having the same form. For, in the equation a x 2n + b x n = c, let x n = y ; then x 2n = if. Substituting, ay 2 +by = c Whence, hy Art. 309, we have or, x • — b± \Jb 2 + kac U — 2a /*n — — b±\Jb 2 + 4tac 2 a from which equation x may he found by extracting the wth root of both members. 314. 1. Solve the equation sc 4 — 5 x 2 = — 4. The equation may be solved as in Art. 313, by representing x- by y. A better method, however, is the following : Completing the square, x* — 5.r 2 + -r- = — 4= + ~r — t Extracting the square root, • x 2 — - = ± - Transposing, a; 2 = 7 r±- = 4orl Whence, x = ± 2 or ±1, Ans. 2. Solve the equation x 6 — 6 x s = 16. Completing the square, x 6 — 6 x a + 9 = 16 + 9 = 25 Extracting the square root, x a — 3 = ± 5 Transposing, cc 3 = 3±5 = 8or — 2 Whence, x = 2 or — \/2, Ans. QUADRATIC EQUATIONS. 229 Here, although the equation is of the sixth degree, we find hut two roots. The equation in reality has six roots, hut this method fails to give more than two. It will he shown here- after how to obtain the other four. 3. Solve the equation x + 4 \Jx = 21. Writing the radical with a fractional exponent, x + 4x^ = 21 which is in the quadratic form. Completing the square, cc + 4^:r + 4 = 21 + 4 = 25 Extracting the square root, \jx + 2 = ± 5 Transposing, y/x = — 2 ± 5 = 3 or — 7 Whence, squaring, x = 9 or 49, Arts. 7 1 4. Solve the equation 3 x 2 + x^ = 3104 x*. Dividing hy x^, 3x% + x% = 3104 which is in the quadratic form. Multiplying hy four times 3, or 12, 36 x% + 12 x% = 37248 Completing the square, 36 x% + 12 x% + 1 = 37249 Extracting the square root, 6 x® + 1 = ± 193 Transposing, 6 x% = — 1 ± 193 = 192 or - 194 5 97 Dividing hy 6, x* = 32 or — y \ /97\i Extracting the fifth root, x b = 2 or — ( jr-J Raising both members to the sixth power, x =64 or (-4)°> Ana. 230 ALGEBRA. EXAMPLES. Solve the following equations : k t 5. .x 4 + 4cc 2 = 117. 11. 3 a;* -^- = -592. 6. a-" 4 - 9 or" 2 + 20 = 0. 12. a 3 -a: 2 =56. 7. a 10 + 31 a- 5 -10 = 22. 13. x-2-s/x = 0. 8. 81 x- + -4 = 82. 14. x% + x* = 756. x- 9. *• + !??? -14 = 60. IS. ^+2 = i=V» a; 2 4 + ya: ya: 3^* 2 10. a; 6 + 20 a; 3 -10 = 59. 16. — — = ^-. x — 5 20 17. Solve the equation (x - 5) 3 - 3 (x - 5) 2 = 40. a 9 9 169 Completing the square, (a; — 5) 3 — 3 (x — 5) 2 + - = 40 + - = — — a 3 ^3 Extracting the square root, (x — 5)* — ■= — ± -=- 3 3 13 Transposing, (a; — 5) ^ = - ± — = 8 or — 5 Squaring both members, (x — 5) 3 = 64 or 25 Extracting the cube root, x — 5 = 4 or \J 25 Whence, x = 9 or 5 + \J 25, QUADRATIC EQUATIONS. 231 Solve the following equations : 18. (x 2 -oa : ) 2 -8(x 2 -5x)=84:. 19. (2 x - 1) 2 - 2 (2 x - 1) = 15. 20. (3 x- -2) 2 - ll(3a; 2 -2) + 10 = 0. 21. (;r J -5) 2 + 29 3 -5)=:96. 22. Solve the equation x* + 10 x 3 + 17 x 2 — 40 aj — 84 = 0. We may write the equation in the form x* + 10 x 3 + 2ox 2 — 8x 2 -A0x = 84 or, (;z 2 +5a;) 2 -8(cr + 5a-)=84 Completing the square, (x 2 +ox) 2 —8(x 2 +5x) + 16 = 100 Extracting the square root, (x 2 + 5 x) — 4 = ± 10 Transposing, (a; 2 + 5 x) = 4 ± 10 = 14 or — 6 Taking the first value, we have x 2 + 5 x = 14 Whence (Art. 309), x = ~ 5 ±^ 5 + 56 = =M? = 2 or - 7. Taking the second value, we have x 2 + 5 x = — 6 w, _5±v/25-24 -5±1 Vv hence, a; = ^ = - = — J or — 3. Ans. x = 2, — 7, — 2, or — 3. Note. In solving equations of this form, our object is to form a perfect trinomial square with the ,r* and x 3 terms, and a portion of the x 2 term. By Art. 305, we may effect this by separating the x 2 term into two parts, one of which shall be the square of the quotient obtained by dividing the x 3 term by twice the square root of the x* term. 232 ALGEBRA. Solve the following equations : 23. x* -12 x 3 + 3±x 2 + 12cc = 35. 24. x i + 2x 3 -2o x- - 26 x + 120 = 0. 25. x 4 - 6 cc 3 - 29 x 2 + 114 x = 80. 26. a 4 + 14 x 3 + 47 x- - 14 a; - 48 = 0. 27. Solve the equation 2 ar + \'2 x 2 + 1 = 11. We may write the equation, (2 x 2 + 1) + y 2 x' 2 + 1 = 12 49 Completing the square, (2 x 2 + 1) + V 2 x 2 + 1 + - = -j- 1 7 Extracting the square root, V^^+l + T^i^ _ 1 7 Transposing, V 2 ^ + * = ~ 2 ± 2 =r3 ° r_4 Squaring, 2 a; 2 + 1 = 9 or 16 Transposing, 2 x 2 = 8 or 15 15 Dividing by 2, a; 2 = 4 or ~^- /15 Whence, x = ± 2 or ± t/ — , ^4?is. Note. In solving equations of this form, add such quantities to hoth members, that the expression without the radical in the first member may be the same as that within, or some multiple of it. Solve the following equations : 28. 2 x 2 + 3 x - 5^2^+3 x + 9 = -3. 29. x- - 6 x + 5 \?x 2 - 6 x + 20 = 46. 30. 4 z 2 + 6 \/4 ar + 12 x - 2 = - 3 (1 + 4 as). 31. x 2 - 10 a: - 2 yV 2 - 10 ^ + 18 + 15 = 0. 32. 3.r 2 +15a:-2v/ar"+5a: + l = 2. QUADRATIC EQUATIONS. 233 XXVII.— SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 315. The most general form of an equation of the second degree containing two unknown quantities, is ax 2 +bxy+cy 2 + dx + ey + f= 0, where the coefficients a, b, c, etc. represent any quantities, positive or negative, integral or fractional. 316. Two equations of the second degree containing two unknown quantities will generally produce, hy elimination, an equation of the fourth degree containing one unknown quantity. Thus, if the equations are x 2 + y = a x + y 2 = b From the first, hy transposition, y = a — x 2 ; substituting in the second, x + (a — x 2 ) 2 = b or, x* — 2 a x 2 + x + a 2 — b = an equation of the fourth degree. The rules for quadratics are, therefore, not sufficient to solve all simultaneous equations of the second degree. In several cases, however, their solution may he effected hy means of the ordinary rules. CASE I. 317. WJien each equation is of the form a x 2 + b y 2 = c. 1. Solve the equations, 3x 2 + Aij 2 = 7Q 3 y 2 - 11 x 2 ^ 4 234 ALGEBRA. Multiplying the first equation by 3, and the second by 4, 9 x 2 + 12 if = 228 12 f-Ux 2 = 16 Subtracting, 53 x 2 = 212 x 2 = ±, x = ±2. Substituting these values in either given equation, When x = 2, y = ± 4. When a? = — 2, y = ±-4. ^4«s. # = 2, ?/ = ± 4 ; or, x = — 2, y = ± 4. EXAMPLES. Solve the following equations : 2. 2 a; 2 + y 2 = 9 ; 5 x 2 + 6 y 2 = 26. 3. 4 a; 2 - 3 f = - 11 ; 11 x 2 + 5 y 2 = 301. 4. 9 x 2 + 24 v/ 2 = 7 ; 72 a; 2 - 180 ?f — - 37. 5. 20 x a - 16 y 2 = 179 ; 5 x 2 - 336 ?/ 2 = 24. CASE II. 318. When one equation is of the first degree. 1. Solve the equations, x 2 + y 2 = 13 x + y — 1 From the second, by transposition, y = 1 — x (1) Substituting in the first, x 2 + 1 — 2 x + x 2 = 13 or, cc 2 — x =6 AVI / A * o no \ • 1 ± S/T+2l 1 ± 5 _ \\ hence (Art. 309), cc = L - = — - — = 3 or — 2. Substituting these values in (1), When a = 3, y = l — 3 — — 2. W T hen x = -2, y = l + 2 = 3. Ans. x = 3, y = — 2 ; or, x = — 2, y = 3. QUADKATIC EQUATIONS. 235 In solving examines under Case II, we find an expression for the value of one of the unknown quantities in terms of the other from the simple equation, which we substitute for that quantity in the other equation, thus producing a quadratic containing only one unknown quantity, by means of which the values of the unknown quantities are readily obtained. Note. Although some examples, in which one equation is of the first degree (Ex. 1 for instance), may be solved by the methods of the next case, yet the method of Case II will be found in general the simplest. EXAMPLES. Solve the following equations : 2. x-\r y= — 1; xy = — 56. . 3. x + y = 3 ; x 2 + y 2 = 29. 4. x s — y 3 = — 37 ; x — y = — 1. 5. x — y = -s--; xy = 20. 6. 10cc + y = 3ccy;3/ — cc = 2. 7. x — y = 5; xy = — 6. 8. x 3 + y 3 = 9 ; x + y = 3. 9. 3x 2 -2icy = 15; 2« + 3y = 12. 10. x - y = 3 ; x 2 + y" = 117. 11. x + y — 11; xy = 18. 12. x — y = 6; x 2 + y* = 90. 13. x 3 + y 3 = 152 ; x + y = 2. 14. x 2 + 3 x y — xf = 23 ; x + 2 y = 7. 15. x 3 -y 3 = 9S; x-y = 2. 16. x + y = -±; x 2 + y 2 = 58. CASE III. 319. When the given equations are symmetrical icith respect to x and y. 236 ALGEBRA. 1. Solve the equations, x" + y 2 = 68 x y = 16 Multiplying the second by 2, 2 x y = 32 Adding this to the first equation, x 2 + 2 x y + y' 2 = 100 (1) Subtracting it from the first equation, x 2 -2xy + y 2 =36 (2) Extracting the square root of (1), x + y = ± 10 (3) Extracting the square root of (2), x — y — ± 6 (4) Equations (3) and (4) furnish four pairs of simple equations, x + y = 10 x + y — 10 x + y = — 10 x + y = — 10 x—y—6 x—y=—Q x—y=6 x—y=—6 2x = 16 2x = 4, 2x = -A 2x = -16 x = 8. x = 2. x = — 2. x = — 8. y = 2. y = 8. y = -S. y = -2. Ans. x = 8, y = 2; x — 2,y = S; x = — 2, y = — 8 ; or, x = — 8, y = — 2. 2. Solve the equations, a;3 + ^ = 133 x 2 — x y + y 2 = 19 Dividing the first equation by the second, x + y = 7 (1) Squaring (1), x 2 + 2 x y + y 2 = 49 (2) Subtracting the second given equation from (2), 3 x y = 30 ; or, 4 x y = 40 (3) Subtracting (3) from (2), x 2 - 2 x y + y 2 = 9 Whence, x — y=±3 (4) QUADRATIC EQUATIONS. 237 Adding (1) and (4), 2 x = 10 or 4 Whence, x = 5 or 2. Substituting these values in (1), When x = 5, y = 2 cc = 2, y = 5. Ans. x — 5, y = 2 ; or, a; = 2, ?/ = 5. The example might have been solved by substituting the value of y derived from (1) in either of the given equations, as in Case II. The student will notice the difference between Examples 1 and 2 as regards the arrangement of the last portion of the work. 3. Solve the equations, x + y = 20 Multiplying the first equation by 2, 2 x 1 + 2 y 2 = 416 (1) Squaring the second equation, x 2 + 2 x y + y 2 = 400 (2) Subtracting (2) from (1), x 2 - 2 x y + y 2 = 16 Whence, x — y = ± 4 (3) Adding the second given equation and (3), 2 x = 24 or 16 Whence, x = 12 or 8. Substituting these values in (3), When » = 12, y = 8 x = 8, y = 12. J??5. aj .= 12, y = 8 ; or, a; = 8, y = 12. This example is solved more readily by the method of Case II; we solve it by Case III merely to show how equations may be solved symmetrically, when one is of the first degree. 238 ALGEBRA. EXAMPLES. Solve the following equations : 4. x' 2 + if = 25- xy = 12. 5. x* + y 2 = S5; xy = 42. 6. x 3 + y 3 = -19; x 2 -xy + y 2 = 19. 7. x 3 - y 3 = - 65 ; x 2 + x y + y 2 = 13. 8. o; + 2/ = l;a;y = — 6. 9. cc 2 + y 2 = 65 ; a; — y = 11. 10. a? 9 +y 2 = 61; x + y = ll. 11. a; 8 — y 8 = 117; a; — y = 3. Note. Exs. 8, 9, 10, and 11 are to be solved like Ex. 3, and not by the method of Case II. In solving Ex. 11, begin by dividing the first equation by the second. CASE IV. 320. When the equations are of the second degree, and homogeneous. Note. Some examples, in which both equations are of the second de- gree and homogeneous, are solved more easily by the methods of Cases I and III, than by that of Case IV. The method of Case IV is to be used only when the example can be solved in no other way. 1. Solve the equations, x 2 — xy = 35 xy + y 2 = 18 Letting y = vx, we have 35 x 2 — v x 2 = 35, or x 2 (1 — v) = 35 ; whence, x 2 = — - - — (1) 18 v x 2 + v 2 x 2 = 18, or x 2 (v + v 2 ) = 18 ; whence, x 2 = - '■ — r v + v QUADRATIC EQUATIONS. 239 Equating the values of x 2 , -z = -= 10 1 — vv+v 2 Clearing of fractions, 35 v + 35 v 2 =18 — 18 v Transposing and uniting, 35 v 2 + 53 v = 18 Whence (Art. 309), - 53 ± V -'809 + 2520 - 53 ± 73 2 9 V = 70 = 70 =7° r -5 2 If v = - , substituting in (1), x 2 = 49, or x = ± 7 Substituting in the equation y = v x, o When x = 7, y=-xT=2 x = -7, !/ = ~x-7 = -2. If v = — ■=, substituting in (1), x 2 = - 7 r- , ov x = ± —j-^ £> 2 y 2 Substituting in the equation y = vx, «n 5 9 5 9 men x =V2' y= ~5 x 72 = ~72 5 9 5 9 ^2'^ 5' N V 2 V 2 .4ns. cc = 7, ?/ = 2; £ = — 7, y = — 2; 5_ 9 j5_ _9_ X ~^/2 ,y ~ S/2 ] ° l,X ~ \J2 ,y ~s/2' Note. In using the equation y~v x, to calculate the value of y when x has been found, care should be taken to use that value of v which n/i* used in getting the particular value of x. EXAMPLES. Solve the following equations : 2. x 2 + xy + ±y 2 = 6; 3x 2 +8y 2 = U. 3. 6x 2 -5xy + 2y 2 = 12; 3 x 2 + 2 x y- 3 y 2 = -3. 240 ALGEBRA. 4. x 2 + x y = 12 ; x y — y 2 = 2. 5. 2 y 2 - 4 x y + 3 x 2 = 17 ; y 2 -x 2 = 16. 6. x 2 + x y — y 2 = 1 ; x 2 — x y + 2 y 2 = 8. 7. 2 x 2 — 2 a; y — ?/ 2 = 3; x 2 + 3a;y + 2/ 2 = 11. 321. We append a few miscellaneous examples, for the solution of which no general rules can be given. Various arti- fices are used ; familiarity with which can only be obtained by experience. 1. Solve the equations, x z — y s = 19 x 2 y — x y 2 = 6 Multiplying the second by 3, 3 x 2 y — 3 x y 2 = 18 (1) Subtracting (1) from the first given equation, x z — 3 x 2 y + 3 x y 2 — y 3 = 1 Extracting the cube root, x — y = 1 (2) Transposing, x = 1 + y (3) Dividing the second given equation by (2), x y = 6 (4) Substituting from (3) in (4), y (1 + y) = 6 or, y 2 + y = 6 m -l±V/T+24 -1±5 Vv hence, ?/ — k — i5 — = ^ or — 3. Substituting in (3), When y = 2, x = 3 y = — 3 } x = — 2. Ans. x = 3, y = 2 ; or, a; = — 2, ?/ = — 3. QUADRATIC EQUATIONS. 241 2. Solve the equations, V * x + y = 12 Let x — u + v, and y = u — v. Then x + y = 2 u ; whence, 2u — 12, or u = 6. From the first given equation, x z + y s = 18 x y Substituting x = 6 + v, and y = 6 — v, we have (6 + vf + (6 - v) 3 = 18 (6 + v) (6 - 1>) Reducing, 432 + 36 v 2 = 648 - 18 v 2 Whence, 54v 2 = 216 v 2 = 4, v = ± 2 Then a5 = 6+v = 6±2 = 8or4. Substituting these values in the second given equation, When x = 8, y = 4 a; = 4, ?/ = 8, ^4ws. 3. Solve the equations, x 2 + y" + x + y = 18 x ?/ = 6 Adding twice the second equation to the first, x 2 + 2 x y + y 2 + x + y — 30 or, (x + y) 2 +(;x + y)=30 N - 1 ± Vl + 120 -1±11 r r Whence, (x + y) = ^ = „ = 5 or — 6. Taking the first value, x + y = 5 (1) and the second given equation, xy = 6 (2) From (1), y=5—x ; substituting in (2), x 2 — 5 x = — 6 5±\/25-24 5±1 Whence, x = —^ = — „ — = 3 or 2. Substituting in (1), When, x = 3, y — 2 cc = 2, y = S. 242 ALGEBRA. Taking the second value, x + y = — 6 (3) and the second given equation, x y = 6 (4) From (3), y — — C — x ; substituting in (4), x 2 + 6 x = — 6 _6±y/36-24 -6+2^3 /0 Whence, a; = ^_ = =r_VL_ — _ 3 ± ^ 3. Substituting in (3), When a= — 3 + v/ 3 ; t/ = — S-—^3. x = -3-sJ3,y = — 3 + sJ3. Ans. x = 3, y = 2 ; cc = 2, ?/ = 3 ; a; = — 3+\/3, y=— 3— ^3; or, x=— 3— y/3, ?/=— 3 + y/3 4. Solve the equations, x* + 7/ = 97 (1) * +y =~1 (2) Raising (2) to the fourth power, x i + 4 x s y + 6 x 2 y 2 + 4 x y 3 + y 4 = 1 (3) Subtracting (1) from (3), 4 a? 3 y + 6 ar ?/ 2 + 4 a; ?/ 3 = — 96 or, 3 a; 2 .?/ 2 + 2 a; ?/ (x 2 + ?/ 2 ) = - 48 (4) But from (2), x 2 + y 2 = l-2xy Substituting in (4), 3 x 2 y 2 + 2 x y (1 - 2 x y) = - 48 or, a: 2 y 2 — 2 cc y = 48 Whence, x y = = — — — = — 6 or 8. Taking the first value, x y = — 6 From (2), y = — 1 — x ; substituting, x 2 + x = 6 Whence, x = j- — = — - = 2 or — 3. Substituting in (2), When x = 2, y = — 3. x = — 3, y = 2. Taking the second value, x y = 8 QUADRATIC EQUATIONS. 243 From (2), y = — 1 — x ; substituting, x 2 + x = — 8 -l±y/T^32 -l±y/-3i \\ hence, x = -= = „ . Substituting in (2), - 1 + V /=7 31 - 1 - V^3l When x = ^ > V~ <j ' - 1 - y/^31 - 1 + \/^31 *= — 2 — ->y=- —2 — Ans. x = 2,y = ~3; x--S,y = 2; x= - 1+ ^~ 31 ^ _1_ S /Z3i _i_y/Z3i _i +v /I73i y = y — i ov > x = 2~ ->y= 2 ' EXAMPLES. Solve the following equations : 5. x + y = 9; \/x+\/y = 3.. 6. x + \J~x~y + y = 19 ; x 2 + x y + f = 133. 7. a; 2 y + ar y 2 = 30 ; x* if + x~y i = 468. 8. x 2 + y 2 — 'x - y = 18 ; x y + x + y = 19. 9. x 2 + 3x + y = 73 - 2 x y ; y 2 + 3 y + x = 44. . « n „ 5 x 1/ xy 10. as» + I f=_£j x - y = -f. ♦ J 4 „, a; 4v/# 33 „ 11. - + -7— = -r ; » — y=& y \jy 4 12 - 2 + 8 = li 5 + S=- 4 13. x 2 y + x y 2 = 30 ; a 3 + y s = 35. 14. cc+V / *'?/=3;?/+V«2/ = — 2 15. x 2 y + y 2 x = 6; - + - = -. 16. r 4 + ?/ 4 = 17; x — y = 3. 17. z 5 - y 5 = - 211 ; cc — y = — 1. 18. ar + y 2 = 7 + as y ; x s + y s = 6 x y — 1. 19. 2x 2 -7x?/-2?/' 2 =:5; 3x y-x 2 + 6 y 2 = U. 244 ALGEBRA. 20. -A^ + Jl^S | + | = 2. y + 3 x + J 2 2 o 21. z + 2 = 7; 2y-3* = -5; a; 2 + y 1 - z~ = 11. 22. xz = y 2 ; (x + y)(z—x—y) = 3; (x + y + z) (z— x— y) = 7. XXVIII. — PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 322. 1. What two quantities are those, the sum of whose squares is 130, and the difference of whose squares is 32 ? Let x = one number, and y = the other. By the conditions, x 2 + y 2 = 130 x 2 -y 2 = 32 Solving these equations, as in Case I, Art. 317, x = 9, y = ± 7 ; # or, x = — 9, y = ± 7. This indicates four answers to the problem : 9 and 7, 9 and — 7, — 9 and 7, — 9 and — 7. Any one of these pairs of values will satisfy the conditions of the problem. 2. A says to B, " The sum of our money is $ 18." B re- plies, "If twice the number of your dollars were multiplied by PROBLEMS. 245 mine, the product would be $ 154." How many dollars had each ? Let x = A's dollars, and y = B's. By the conditions, x + y = 18 2 x y = 154 Solving these equations, as in Case II, Art. 318, x = li y = U; or, x = 11, y = 7. That is, either Alias $7, and B $11, or A has $11, and B $7. 3. The price of two coats and one vest is $ 38. And the price of a coat less that of a vest, is to $ 23, as $ 7 is to the sum of the prices of a coat and vest. What is the price of a coat, and what of a vest ? Let x = the price of a coat in dollars, and y = the price of a vest. By the conditions, 2 x + y = 38 and x — y : 23 = 7 : x + y or (Art. 181), a 2 -?/ 2 = 161 Solving these equations, as in Case II, Art. 318, x = 15, y = 8 ; 107 100 x = — ,y = - — . Only the first answer is admissible, as a negative value of either unknown quantity does not answer to the conditions of the problem. Hence, the price of a coat is $15, and of a vest, $ 8. Note. The note after Ex. 3, Art. 311, applies with equal force to the problems in this chapter. 246 ALGEBRA. PROBLEMS. 4. The difference of two quantities is 5, and the sum of their squares is 193. What are the quantities ? 5. There are two quantities whose product is 77, and the difference of whose squares is to the square of their difference as 9 to 2. Required the quantities. 6. A and B have each a field, in the shape of an exact square, and it requires 200 rods of fence to enclose hoth. The contents of these fields are 1300 square rods. What is the value of each at $ 2.25 per square rod ? 7. Two gentlemen, A and B, were speaking of their ages. A said that the product of their ages was 750. B replied, that if his age were increased 7 years, and A's were diminished 2 years, their product would be 851. Required their ages. 8. A certain garden is a rectangle, and contains 15,000 square yards, exclusive of a walk, 7 yards wide, which sur- rounds it, and contains 3696 square yards. Required the length and breadth of the garden. 9. What k two numbers are those whose difference multi- plied by the less produces 42, and by their sum, 133 ? 10. A and B lay out money on speculation. The amount of A's stock and gain is $ 27, and he gains as much per cent on his stock as B lays out. B's gain is $ 32 ; and it appears that A gains twice as much per cent as B. Required the capi- tal of each. 11. I bought sugar at such a rate, that the price of a pound was to the number of pounds as 4 to 5. If the cost of the whole had been 45 cents more, the number of pounds would have been to the price of a pound as 4 to 5. How many pounds were bought, and what was the price per pound ? 12. A and B engage in speculation. A disposes of his share for $ 11, and gains as many per cent as B invested dollars. PROBLEMS. 247 B's gain was $ 36, and the gain upon A's investment was 4 times as many per cent as upon B's. How much did each invest ? 13. A man bought 10 ducks and 12 turkeys for $ 22.50. He bought 4 more ducks for $ 6, than turkeys for $ 5. What was the price of each ? 14. A man purchased a farm in the form of a rectangle, whose length was 4 times its breadth. It cost £ as many dol- lars per acre as the field was rods in length, and the number of dollars paid for the farm was 4 times the number of rods round it. Required the price of the farm, and its length and breadth. 15. I have two cubic blocks of marble, whose united lengths are 20 inches, and contents 2240 cubic inches. Required the surface of each. 16. A's and B's shares in a speculation altogether amount to $ 500. They sell out at par, A at the end of 2 years, B of 8, and each receives in capital and profits $ 297. How much did each embark ? 17. A person has $ 1300, which he divides into two portions, and loans at different rates of interest, so that the two por- tions produce equal returns. Jf the first portion had been loaned at the second rate of interest, it would have produced $ 36 ; and if the second portion had been loaned at the first rate of interest, it would have produced $49. Required the rates of interest. 18. Two men, A and B, bought a farm of 104 acres, for which they paid $320 each. On dividing the land, A says to B, "If you will let me have my portion in the situation which I shall choose, you shall have so much more land than I, that mine shall cost $ 3 per acre more than yours." B accepted the proposal. How much land did each have, and what was the price of each per acre ? 248 ALGEBRA. 19. A and B start at the same time from two distant towns. At the end of 7 daj's, A is nearer to the half-way house than B is, by 5 miles more than A's day's journey. At the end of 10 days they have passed the half-way house, and are distant from each other 100 miles. Now it will take B 3 days longer to perform the whole journey than it will A. Required the distance of the towns, and the rate of walking of A and B. 20. Divide the number 4 into two such parts that the prod- uct of their squares shall be 9. 21. The fore-wheel of a carriage makes 15 revolutions more than the hind-wheel in going 180 yards ; but if the circumfer- ence of each wheel were increased by 3 feet, the fore-wheel would only make 9 revolutions more than the hind-wheel in going the same distance. Find the circumference of each wheel. 22. A ladder, whose foot rests in a given position, just reaches a window on one side of a street, and when turned about its foot, just reaches a window on the other side. If the two positions of the ladder are at right angles to each other, and the heights of the windows are 36 and 27 feet re- spectively, find the width of the street and the length of the ladder. 23. A and B engaged to reap a field for 90 shillings. A could reap it in 9 days, and they promised to complete it in 5 days. They found, however, that they were obliged to call in C, an inferior workman, to assist them the last two days, in consequence of which B received 3.?. 9d. less than he other- wise would have done. In what time could B and C each reap the field? 24. Cloth, being wetted, shrinks J in its length and ,\ ; in its- width. If the surface of a piece of cloth is diminished by 5^ square yards, and the length of the four sides by 4} yards, what was the length and width of the doth originally? THEORY OF QUADRATIC EQUATIONS. 249 XXIX. — THEORY OF QUADRATIC EQUA- TIONS. 323. A quadratic equation cannot have more than two roots. We have seen (Art. 304) that every complete quadratic equation can be reduced to the form x 2 + p x = q. Suppose, if possible, that a quadratic equation can have three roots, and that r 1; r 2 , and r 3 are the roots of the equation x°- + p x — q. Then (Art. 166), r 1 2 +pr 1 = q (1) r 2 2 +pr 2 — q (2) r 3 2 +pr 3 — q (3) Subtracting (2) from (1), (?y — r.F) + p (r x — r 2 ) = Dividing through by r y — r 2 , which by supposition is not zero, as the roots are not equal, r \ + r 2 +p = Similarly, by subtracting (3) from (1), we have n + n + p — o Hence, r x + r 2 + p = r x + r s -f p or, r 2 = r 3 . That is, two of the roots are identical. Therefore, a quad- ratic equation cannot have more than two roots. DISCUSSION OF THE GENERAL EQUATION. 324. By Art. 305, the roots of the equation x 2 + p x = q are -p-\- \/p 2 + ±q 1 —p — \Jp 2 +4:q 250 ALGEBRA. 1. Suppose q positive. Since p 2 is essentially positive (Art. 227), the quantity under the radical sign is positive and greater than ,p° ; so that the value of the radical is greater than P- Hence, one root is positive, and the other negatn r e. If p is positive, the negative root is numerically the larger; if p is zero, the roots are numerically equal ; and if p is nega- tive, the positive root is numerically the larger. 2. Suppose q equal to zero. The quantity under the radical sign is now equal to p 2 ; so that the value of the radical is p. Hence, one of the roots is equal to 0. The other root is positive when p is negative, and negative when p is positive. 3. Suppose q negative, and 4 q < p 2 . The quantity under the radical sign is now positive and less than p> 2 ; so that the value of the radical is less than p. If p is positive, hoth roots are negative ; and if p> is nega- tive, both roots are positive. 4. Suppose q negative, and 4 q =p 2 . The quantity under the radical sign is now equal to zero ; so that the two roots are equal ; being positive if p is negative, and negative if p is positive. 5. Suppose q negative, and 4 q > p 2 . The quantity under the radical sign is now negative ; hence, by Art. 282, both roots are imaginary. 325. All these cases may be readily verified by examples. Thus, in the equation x 2 — 3;r = 70, as p is negative and q positive, we should expect to find one root positive and the other negative, and the positive root numerically the larger And this is actually the rase, for on solving the equation, wfc find x = 10 or - 7. THEORY OF QUADRATIC EQUATIONS. 251 326. From the quadratic equation x 2 +px = q, denoting the roots by r x and r. 2 , we have -p+S/Y + lq _ 1m -p-Sjp' + lq 2 n = — ^ , and r 2 Adding these together, we have 2p r x + r 2 = — -^- = —p. Multiplying them together, we have n r 2 = £=M±±1± (Art. 106) = - *± = - q . That is, if a quadratic equation be reduced to the form x 2 + p x = q, the algebraic sum of the roots is equal to the co- efficient of the second term, with its sign changed • and the product of the roots is equal to the second member, with its sign changed. 327. The equation a x 2 + b x + c = 0, by transposing c, and dividing each term by a, becomes x 2 -\ bx c a a Denoting the roots of the equation by x l and x 2 , we have, by the previous article, b _c Xi -j- Xo — , and x± x.y — — • a a 328. A Quadratic Expression is a trinomial expression of the form a x 2 + b x + c. The principles of the preceding article enable us to resolve any quadratic expression into two binomial factors. The expression a x' 2 + b x + c may be written f bx c a I x -\ 1 — V a a 252 ALGEBRA. b c By the previous article, - = —(% + x 2 ), and - =x 1 x 2 , where x x and x 2 are the roots of the equation ax 2j rbx + c = Q; which, we ohserve, may he obtained by placing the given expression equal to 0. Hence, ax 2 + bx + c = a [x 2 — (x x + x 2 ) x + x x x 2 ~\. The expression in the bracket may be written /V»** rtn /yt ry* /y> I /y» sy which, by Case II, Chap. VIII, is equal to (x — x^ (x — x 2 ). Therefore, a x 2 + b x + c = a (x — x x ) (x — x 2 ). 1. Factor 6 x 2 + 11 x -+- 3. Placing the expression equal to 0, and solving the equation thus formed, we find _ - 11 + ^121 - 72 _ - 11 ± 7 _ 3 1 X ~~ ~V1~ ~ _ ~~12 - ~2' 0r ~3- Then, a = 6, x 1 = — ^, x 2 = — 5. Therefore, 6 x 2 + 11 x + 3 = 6 (x + |) (» + 5) = (2 a; + 3) (3 x + 1), Ans. 2. Factor 4 + 13 x - 12 x 2 . Placing the expression equal to 0, and solving the equation formed, we have _ - 13 ± y/ 169 + 192 _ - 13 ± 19 _ 4 1 X ~ -24 -24 ~3' ° r 4 4 1 Then, a = — 12,x 1 = 7 r, x 2 = — -. THEORY OF QUADRATIC EQUATIONS. 253 Therefore, 4 + 13 x - 12 x 2 = - 12 he - ^J (a; + ^J = -3{x-l)4(x + \) — (4 — 3 x) (4 a; + 1), ^4?is. Note. It should be remembered, in using the formula a (x-x{) (x - a'2 N , that a represents the coefficient of x 2 in the given expression J hence, in Example 2, we made a= - 12. EXAMPLES. Factor the following expressions : 3. x 1 + 73 x + 780. 9. 8z 2 + 18z-5. 4. x 2 - 11 x + 18. 10. 4 z 2 - 15 * + 9. 5. x 2 -4cc-60. 11§ 2x 2 +x-6. 6. a 8 + 10 a; -39. 12. 9x 2 -12a- + l. 7. 2 a; 2 -7 a; -15. 13. l-8x-x 2 . 8. 21 a- 2 + 58 a; + 21. 14. 49 x 2 + 14 a- - 19. 329. The principles of Art. 328 furnish a method of form- ing a quadratic equation which shall have any required roots. For, the equation a x 2 + b x + c = 0, if its roots be denoted by x x and x 2 , may be written, by Art. 328, a (x — Xy) (x — x 2 ) = 0, or (x — £Cj) (x — a- 2 ) = 9. Hence, to form an equation whose roots shall be x v and x.,, we subtract each of the two roots from x, and place the product of the resulting binomials equal to zero. 7 1. Required the equation whose roots are 4 and — -r • By the rule, (x — 4) (x + -)= 254 or, ALGEBRA. x 2 - 9x -7 = = 4x 2 - 9x- -28 = = o, Ans. Clearing of fractions, EXAMPLES. Form the equations whose roots are 17 2. 1 and — 2. 5. 7 and — 6 \ . 8. — -^- and 0. 8 4 3. 4 and 5. 6. — - and - . 9. 1 + sj 5 and 1 — ^5. o i 4. 3 and —■=. 7. — 2 J and — 3^. 10. m + \/ w and m — \Jn. o 330. By Art. 328, the equation a x 2 + b x + c = may be written (x — a^) (x — x 2 ) = 0, if acj and x 2 are its roots;- we observe that the roots may be obtained by placing the factors of the first member separately equal to zero, and solving the simple equations thus formed. This principle is often useful in solving equations. 1. Solve the equation (2x — 3) (3 x + 5) = 0. 3 Placing the first factor equal to zero, 2x — 3 = 0, or a; = - . A 5 Placing the second factor equal to zero, ox+ 5=0, orx= — - . A 3 5 Ans. x = - or — ^ . A o 2. Solve the equation x' 2 + 5 x = 0. The equation may be written x (x + 5) = Placing the first factor equal to zero, x = 0. Placing the second factor equal to zero, x + 5 = 0, or x = — 5. Ans. x = or — 5. THEORY OF QUADRATIC EQUATIONS. 255 EXAMPLES. Solve the following equations : 3. (as-|) (as-2) = 0. 9. 2as 8 -18as = 0. 4. (as+.5)(as-l) = 0. 10. (2as + 5)(3as-l) = 0. 5. (as-?) (as + ?)=0. 11. (aas + b) (cx-d) =0. 6. (as + 8)(as + i)=0. 12. (x 2 -4) (as 2 -9) = 0. 7. 2as 2 -13as = 0. 13. (3 a; + 1) (4 x* - 25) = 0. 8. 3 as 3 + 12 a: 2 = 0. 14. (as 2 -a)(as 2 -aas-£)=0. 15. as (2 as + 5) (3 x -7) (4 x + 1) = 0. 16. (x 2 -5x + 6)(x 2 + 7x+ 12) (2 x* + 9x-5) = 0. 331. Many expressions may be factored by the artifice of completing the square, used in connection with the method of Case IV, Chapter VIII. 1. Factor x i + a\ x* + a 4 = x* + 2 x 2 a 2 +a*-2 x 2 a 2 = (x 2 + a 2 ) 2 - (a x ^/ 2) 2 = (Art. 117) (x 2 + a x ^ 2 + a 2 ) (x 2 — a x \J 2 + a 2 ), Ans. 2. Factor x 2 — ax + a 2 . as 2 — ax + a 2 = x 2 + 2ax + a 2 — 3ax. = (x + a) 2 -(^3axY = (x + V^3 a x + a) (x — \J3 a x + a), Ans. 256 ALGEBRA. EXAMPLES. Factor the following expressions : 3. x 2 + l. 5. a 2 -3ab + b*. 7. x 2 -x-l. 4. x 2 +x+l. 6. x i -lx 2 y 2 -^y\ 8. m i + m 2 n 2 +?i\ 332. We have seen (Art. 330) that any equation whose first member can be factored, and whose second member is zero, may be solved by placing the factors separately equal to zero and solving the equations thus formed. This method of solution is frequently the only one which will give all the roots of the equation. 1. Solve the equation x 3 = 1. The equation may be written x 3 — 1 = 0, or (Art. 119), (x - 1) (x 2 + x + 1) = 0. Placing the first factor equal to zero, x — 1 = 0, or x = 1. Placing the second factor equal to zero, x 2 + x + 1 = 0, or x 2 + x = — 1 Whence (Art, 309), - - = ^ ± ^ x = 2 2 Hence, x = 1 or ^1 , Ans. EXAMPLES. Solve the following equations : 2. x i = -l. 4. x 4 +« 4 = 0. 6. x 6 = l. 3. x 3 = -l. 5. X 4- X * + 1 = 0. 7. rr 4 - — +1 = 0. These examples afford an illustration of the statement made in Art. 167 that the degree of an equation indicates the num- ber of its roots. DISCUSSION OF PROBLEMS. 257 XXX. — DISCUSSION OF PROBLEMS LEADING TO QUADRATIC EQUATIONS. 333. In the discussion of problems leading to quadratic equations, we find involved the same general principles which have been established in connection with simple equations (Arts. 205-212), but with certain peculiarities. These peculiarities will be now considered. They arise from two facts : 1. That every quadratic equation has two roots • and 2. That these roots are sometimes imaginary. 334. In the solution of problems involving quadratics, it has been observed that the positive root of the equation is usually the true answer ; and that, when both roots are posi- tive, there may be two answers, either of which conforms to the given conditions. The reason why results are sometimes obtained which do not apply to the problem under consideration, and are there- fore not admissible, is that the algebraic mode of expression is more general than ordinary language ; and thus the equa- tion which conforms properly to the conditions of the problem will also apply to other conditions. 1. Find a number such that twice its square added to three times the number may be 65. Let x = the number. Then 2 x 2 + 3 x = 65 (1) 13 Whence, x = 5 or — . The positive value alone gives a solution to the problem in the sense in which it is proposed. To interpret the negative value, we observe that if we change x to — x, in equation (1), the term 3 x, only, changes 258 ALGEBRA. its sign, giving as a result the equation 2 x' 2 — 3 x = 65. Solv- 13 ing this equation, we shall find x = — or — 5, which values only differ from the others in their signs. We therefore may 13 consider the negative solution, — , taken independently of Li its sign, the proper answer to the analogous problem (Art. 205): " Find a number such that twice its square diminished by three times the number may be Go." 2. A farmer bought some sheep for $ 72, and found that if he had bought 6 more for the same money, he would have paid $ 1 less for each. How many sheep did he buy ? Let x = the number of sheep bought. 72 Then — = the price paid for one, x 72 and = the price paid, if 6 more. JO \~ O 72 72 By the conditions, — = - + 1 J ' x x + 6 Whence, x = 18 or — 24. Here the negative result is not admissible as a solution of the problem in its present form; the number of sheep, there- fore, was 18. If, in the given problem, "6 more " be changed to "6 fewer" and "$1 less" to "$1 more," 24 will be the true answer. Hence, we infer that A negative result, obtained as one of the answers to a prob- lem, is sometimes the answer to another analogous problem, formed by attributing to the unknown quantity a quality directly opposite to that which has been attributed to it. DISCUSSION OF PROBLEMS. 259 INTERPRETATION OF IMAGINARY RESULTS. 335. It lias been shown (Art. 324) under what circum- stances a quadratic equation will be in form to produce imagi- nary roots. It is now proposed to interpret such results. Let it be required to divide 10 into two such parts that their product shall be 26. Let x = one of the parts. Then 10 — x = the other. By the conditions, x (10 — x) = 26 Whence, x = 5 ± y/— 1. Thus, we obtain an imaginary result. We therefore con- clude that the problem cannot be solved numerically ; in fact, if we call one of the parts 5 + y, the other must be 5 — y, and their product will be 25 — y' 2 . which, so long as y is numerical, is less than 25. But we are required to find two numbers whose sum is 10 and product 26 ; there are, then, no such numbers. Had it been required to find two expressions, whose sum is 10 and product 25. the answer 5 + \/ — 1 and 5 — \j— 1 would have satisfied the conditions. The given problem, however, expresses conditions incom- patible with each other, and, consequently, is impossible. Hence, Imaginary results indicate that the problem is impossible. PROBLEM OF THE LIGHTS. 336. The principles of interpretation will be further illus- trated in the discussion of the following general problem. Find upon the line which joins two lights, A and B, the point which is equally illuminated by them : admitting that the intensity of a light, at a given distance, is equal to its 2G0 ALGEBRA. intensity at the distance 1, divided by the square of the given distance. C" A C B C — I 1 J 1 1 — Assume A as the origin of distances, and regard all dis- tances estimated to the right as positive. Let a denote the intensity of the light A, at the distance 1 ; b the intensity of the light B, at the distance 1 ; and c the distance A B, between the two lights. Suppose C the point of equal illumination, and let x repre- sent the distance from it to A, or the distance AC. Then, c — x will represent the distance B C. By the conditions of the problem, since the intensity of the light A, at the distance 1, is a, at the distance x it is — = : and X' since the intensity of the light B, at the distance 1, is b, at the distance c — x it is -. ^ . But, by supposition, at C these intensities are equal ; hence, a b Whence, x 2 ' (c - xf ' c—x ~ ( c - x) 2 b or X 2 a _ + v* x \l a From this equation we obtain as the values of x : c\J a i \J a x or, \Ja + ^b w \\ja + ^by c\J a / sj a \ \J a — \J b \\J a — \J b)' Since both a and b are positive, the two values of x are both real. Hence, There are two points of equal illumination on the line of the lights. DISCUSSION OF PROBLEMS. 261 Since there are two lights, c must always he greater than ; consequently neither a, b, nor c can he 0. The problem, then, admits properly of only these three different suppositions : 1. a > b. 2. a < b. 3. a=b. We shall now discuss the values of x under each of these suppositions. 1. a > b. In this case, the first value of x is less than c ; because - being a proper fraction, is less than 1. This value y/ a + \J b c of x is also greater than - ; because, the denominator being less than twice the numerator, as b is less than a, the fraction is greater than \. Hence, the first point of equal illumination is at C, between the two lights, but nearer the lesser one. The second value of x is greater than c : because —A- — , ya — \J b being an improper fraction, is greater than 1. Hence, the second point is at C, in the prolongation of the line A B, be- yond the lesser light. These results agree with the supposition. For, if a is greater than b, then B evidently is the lesser light. Hence, both points of equal illumination will be nearer B than A ; and since the two lights emit rays in all directions, one of the points must he in the prolongation of A B beyond both lights. 2. a < b. In this case, the first value of x is positive. It is also less than C -\ because . ^ C ' . , , having the denominator greater 2 v a + y b than twice the numerator, b being greater than a, is less than h- Hence, the first point of equal illumination is between the lights, but nearer A, the lesser light, The second value of x is negative, because the denominator y/ a — y 1 b is negative ; which must he interpreted as measur- 262 ALGEBKA. ing distance from A towards the left (Art. 205). Hence, the second point of equal illumination is at C", in the prolongation of the line, at the left of the lesser light, A. These results correspond with the supposition; the case being the same as the preceding one, except that A is now the lesser light. 3. a = b. In this case, the first value of x is positive, and equal to ^ . Li Hence, the first point of equal illumination is midway be- tween the two lights. The second value of x is not finite: because -; — p- , if V « — V b I a = b, reduces to ~- = cc (Art. 210), which indicates that no finite value can be assigned to x. Hence, there is no second point of equal illumination in the line A B, or its prolongation. These results agree with the supposition. For, since the lights are of equal intensity, a point of equal illumination will obviously be midway between them ; and it is evident that there can be no other like point in their line. The preceding discussion illustrates the precision with which algebraic processes will conform to every allowable interpreta- tion of the enunciation of a problem. XXXI. — RATIO AND PROPORTION. 337. The Ratio of one quantity to another of the same kind is the quotient arising from dividing the first quantity by the second (Art. 181). Thus, the ratio of a to b is - , or a : b. b EATIO AND PROPORTION. 263 338. The Terms of a ratio are the two quantities required to form it. Of these, the first is called the antecedent, and the second the consequent. Thus, in the ratio a : b, a and b are the terms, a the ante- cedent, and b the consequent. 339. A Proportion is an equality of ratios (Art. 181). Thus, if the ratios a : b and c : d are equal, they form a pro- portion, which may he written a : b = c : d, or a : b : : c : d. 340. The Terms of a proportion are the four terms of its two ratios. The first and third terms are called the antece- dents ; the second and fourth, the consequents; the first and last, the extremes ; the second and third, the means ; and the terms of each ratio constitute a couplet. Thus, in a : b = c : d, a and c are antecedents ; b and d, con- sequents ; a and d, extremes; b and c, means; a and b, the first couplet ; and c and d, the second couplet. 341. A Proportional is any one of the terms of a propor- tion ; a Mean Proportional between two quantities is either of the two means, when they are equal ; a Third Proportional to two quantities is the fourth term of a proportion, in which the first term is the first of the quantities, and the second and third terms each equal to the second quantity ; a Fourth Pro- portional to three quantities is the fourth term of a proportion whose first, second, and third terms are the three quantities taken in their order. Tims, in a : b = b : e, b is a mean proportional between a and c ; and c is a third proportional to a and b. In a : b — c : d, d is a fourth proportional to a, b, and c. 342. A Continued Proportion is one in which each conse- quent is the same as the next antecedent ; as, a : b = b : c = c : d =■ d : e. 264 ALGEBRA. PROPERTIES OF PROPORTIONS. 343. Wlien four quantities are in proportion-, the prodwt of the extremes is equal to the product of the means. Let the proportion be a : b = c : d. This may he written (Art. 337), a c b = d Whence, ad = b c. Hence, if three quantities be in continued proportion, the product of the extremes is equal to the square of the means. Thus, if a : b = b : c then, a c = b 2 . By this theorem, if three terms of a proportion are given, the fourth may be found. Thus, if then, Whence, 344. If the product of two quantities be equal to the prod- uct of two others, one pair of them may be made the extremes, and the other pair the means, of a proportion. Thus, if a d = b c ad be a c Dividing bv b d, -=— =■ = rr— ; , or T = -= & J bd bd' b d Whence (Art. 337), a:b = c:d. a : b = c : x a x = bc x _bc a RATIO AND PROPORTION. 265 In a similar manner, we might derive from the equation a d = b c, the following proportions : a : c = b : c 1, b : d = a : c, c : d = a : b, d : b = c : a, etc. 345. If four quantities are in proportion, they will be in proportion by Alternation; that is, the antecedents will have to each other the same ratio as the consequents. Thus, if a : b = c : d then (Art. 343), ad = bc Whence (Art. 344), a : c = b : d. 346. If four quantities are in proportion, they will be in proportion by Inversion; that is, the second term will be to the first, as the fourth is to the third. Thus, if a : b — c : d then, ad — be Whence, b : a = d : c. 347. If four quantities are in proportion, they will be in proportion by Composition; that is, the sum of the first two terms ivill be to the first term, as the sum of the last two terms is to the third term. Thus, if a : b — c : d then, ad = b c Adding hoth members to a c, ac + ad = ac + bc, or a (c ' + d) = c {a + b) Whence, a + b: a = c + d: c (Art. 344). Similarly, we may show that a + b : b — c + d : d. 266 ALGEBRA. 348. If four quantities are in proportion, they will be in proportion by Division; that is,, the difference of the first two terms will be to the first term, as the difference of the last two terms is to the third term. Thus, if a : b = c : d then, a d = b c Subtracting both members from a c, ac — ad = ac — be, or a (c — d)=c (a — b) Whence, a — b : a = c — d : c. Similarly, we may prove that a — b : b = c — d : d. 349. If four quantities are in proportion, they will be in proportion by Composition and Division; that is, the sum of the first two terms will be to their difference, as the sum of the last two terms is to their difference. (1) (2) Thus, if a : b — c : d by Art. 347, a + b c + d a e and, by Art. 348, a — b c — d a c Dividing (1) by (2), a + b c + d a — b c — d or, a -f- b : a — b = c + d : c d. 350. Quantifies which are proportional to the same quan- tities, are propjortional to each other. Thus, if a :b = e:f and c : d = e :f . ae.ee then, r = 3 and - 7 = ^ Therefore, -=- b d Whence, a:b = c:d. RATIO AND PROPORTION. 267 351. If any number of quantities are proportional, any antecedent is to its consequent, as the sum of all the antece- dents is to the sum of all the consequents. Thus, if a : b = c : d = e :f then (Art. 343), ad = b c and af-=be also, a b = « b Adding, a (b + d +/) = b (a + c + e) Whence (Art. 344), a :b =a + c + e : b + d +f 352. If four quantities are in proportion, if the first and second be multiplied or divided by any quantity, as also the third and fourth, the resulting quantities will be in proportion* Thus, if a : b = c : d then, a c b = d Therefore, ma nc m b n d Whence, m a : mb = n c : n d. In a similar manner we could prove a b c d m ' m n' n' Either m or n may be made equal to unity. That is, either couplet may be multiplied or divided, without multiplying or dividing the other. 353. If four quantities are in proportion, and the first ami third be multiplied or divided by any quantity, as also the second and fourth, the resulting quantities will be in jiro- portion. 268 ALGEBRA. Thus, if a : b = c : d then, Therefore, a c b = d m a m c nb n d Whence, m a : nb = m e : nd. In a similar manner we could prove a b _ c d m n m ' n Either m or n may he made equal to unity. 354. If there be two sets of proportional quantities, the products of the corresponding terms will be in proportion. Thus, if a :b = c : d and e:f=g:h a c .. e q then, - = - and -=- Therefore, b d f h ae eg bf dh Whence, ae :bf= c g : d h. 355. If four quantities are in proportion, like poioers or like roots of these quantities will be in proportion. Thus, if a :b = c :d then, r = --,', therefore, — = — b d ' b n d n Whence, a n :b n = c n : d n . In a similar manner we could prove y/ a : y/ b = y/ c : y d. 356. If three quantities are in continued proportion, the first is to the third, as the square of the first is f>> the square of the second. RATIO AND PROPORTION. 269 Thus, if a:b = b : c a b then, a b c a 2 aba Multiplying by -, y=l* c c Whence, a: c — a~ : b~. In a similar manner we could prove that if a : b = b : c = c : d, then a\d=-a z : b 3 . Note. The ratio a 2 : b 2 is called the duplicate ratio, and the ratio a? : b 3 the triplicate ratio, of a : b. PROBLEMS. 357. 1. The last three terms of a proportion being 18, 6, and 27, what is the first term ? 2. The first, third, and fourth terms of a proportion being 4, 20, and 55, respectively, what is the second term ? 3. Find a fourth proportional to \, \, and \. 4. Find a third proportional to 5 and 3. 5. Find a mean proportional between 2 and 8. 6. Find a mean proportional between 6 and 24. 7. Find a mean proportional between 49 and 4. 8. Find two numbers in tbe ratio of 2\ to 2, such that when each is diminished by 5, they shall be in the ratio of lj to 1. 9. Divide 50 into two such parts that the greater increased by 3, shall be to the less diminished by 3, as 3 to 2. 10. Divide 27 into two such parts that their product shall be to the sum of their squares as 20 to 41. 11. There are two numbers which are to each other as 4 to 9, and 12 is a mean proportional between them. What are the numbers ? 270 ALGEBRA. 12. The sum of two numbers is to their difference as 10 to 3, and their product is 364. What are the numbers- ? 13. Find two numbers such that if 3 be added to each, they will be in the ratio of 4 to 3 ; and if 8 be subtracted from each, they will be in the ratio of 9 to 4. 14. There are two numbers whose product is 96, and the difference of their cubes is to the cube of their difference as 19 to 1. What are the numbers ? 15. Each of two vessels contains a mixture of wine and water ; a mixture consisting of equal measures from the two vessels, contains as much wine as water; and another mixture consisting of four measures from the first vessel and one from the second, is composed of wine and water in the ratio of 2 to 3. Find the ratio of wine to water in each vessel. 16. If the increase in the number of male and female criminals be 1.8 per cent, while the decrease in the number of males alone is 4.6 per cent, and the increase in the number of females alone is 9.8 per cent, compare the number of male and female criminals, respectively, at the first time mentioned. 17. A railway passenger observes that a train passes him, moving in the opposite direction, in 2 seconds ; whereas, if it had been moving in the same direction with him, it would have passed him in 30 seconds. Compare the rates of the two trains. XXXII. — VARIATION. 358. Variation, or general proportion, is an abridged method of expressing common proportion. Thus, if A and B be two sums of money loaned for equal times, at the same rate of interest, then A : B = A's interest : B's interest VARIATION. 271 or, in an abridged form, by expressing only two terms, the in- terest varies as the principal ; thus (Art. 23), The interest oc the principal. 359. One quantity varies directly as another, when the two increase or decrease together in the same ratio. Sometimes, for the sake of brevity, we say simply one quan- tity varies as another, omitting the word " directly/' Thus, if a man works for a certain sum per day, the amount of his wages varies as the number of days during which he works. For, as the number of days increases or decreases, the amount of his wages will increase or decrease, and in the same ratio. 360. One quantity varies inversely as another, when the first varies as the reciprocal of the second. Thus, if a courier has a fixed route, the time in which he will pass over it varies inversely as his speed. That is, if he double his speed, he will go in half the time; and so on. 361. One quantity varies as two others jointly, when it has a constant ratio to the product of the other two. Thus, the wages of a workman will vary as the number -of days he has worked, and the wages per day, jointly. 362. One quantity varies directly as a second and inversely as a third, when it varies jointly as the second and the recip- rocal of the third. Thus, in physics, the attraction of a planetary body varies directly as the quantity of matter, and inversely as the square of the distance. 363. If A varies as B, then A is equal to B multiplied by some constant quantity. Let a and b denote one pair of corresponding values of two quantities, and A and B any other pair. Then, by Art. 358, 272 ALGEBRA. A:a = B:b Whence (Art. 343), Ab = a B, or A=^B • a Denoting the constant ratio - by m, A = m B. Hence, also, when any quantity varies as another, if any two pairs of values of the quantities be taken, the four will be proportional. For, if A oc B, and. A' and B' be any pair of values of A and B, and A" and B" any other pair, by Art. 363, A' = m B', and A" = m B" a A 1 A Whence, -~ = m, and — - Therefore, A" B 1 B" or (Art. 337), A':B' = A" : B 'ii 364. The terms used in Variation may now be distin- guished as follows : 1. If A = m B, A varies directly as B. Ml 2. If A = — , A varies inversely as B. 3. If A = m B C, A varies jointly as B and C. 4. If A — , A varies directly as B, and inversely as C. Problems in variation, in general, arc readily solved by con- verting the variation into an equation, by the aid of Art. 364. VAKIATION. 273 EXAMPLES. 365. 1. Given that y oc x, and when x = 2, y = 10. Re- quired the value of y in terms of x. If y oc x, by Art. 364, y = mx Substituting x = 2 and y = 10, 10 = 2 m, whence m = 5. Hence, the required value is y = 5 a?. 2. Given that 7/ oc »■, and that v/ = 2 when x = l. What will he the value of y when x = 2 ? 3. If y oc £, and y = 21 when £ = 3, find the value of y in terms of z. 4. If x varies inversely as y, and x = 4 when y = 2, what is the value of cc when y = 6? 5. Given that z varies jointly as x and y, and that z = 1 when x = 1 and ?/ = 1. Find the value of z when x = 2 and y = 2. 6. If ?/ equals the sum of two quantities, of which one is constant, and the other varies as x y ; and when x = 2, y = — 2 J-, hut when x = — 2, y = 1 ; express y in terms of x. 7. Two circular plates of gold, each an inch thick, the diam- eters of which are 6 inches and 8 inches, respectively, are melted and formed into a single circular plate one inch thick- rind its diameter, having given that the area of a circle varies as the square of its diameter. 8. Given that the illumination from a source of light varies inversely as the square of the distance ; how much farther from a candle must a book, which is now 3 inches off, be removed, so as to receive just half as much light ? ' 9. A locomotive engine without a train can go 24 miles an hour, and its speed is diminished by a quantity which varies as the square root of the number of cars attached. With four cars its speed is 20 miles an hour. Find the greatest number of cars which the engine can move. 274 ALGEBEA. XXXIII. — ARITHMETICAL PROGRESSION. 366. An Arithmetical Progression is a series of quanti- ties, in which each term is derived from the preceding term by adding a constant quantity, called the com raon difference. 367. When the series is increasing, as, for example, 1,3,5,7,9,11, each term is derived from the preceding term by adding a positive quantity; consequently the common difference is positive. When the series is decreasing, as, for example, 19, 17,15,13,11,9, each term is derived from the preceding term by adding a negative quantity; consequently the common difference is negative. 368. Given the first term, a, the common difference, d, and the number of terms, n, to find the last term, I. The progression will be a, a + d, a + 2 d, a + 3 d, We observe that these terms differ only in the coefficient of d, which is 1 in the second term, 2 in the third term, 3 in the fourth term, etc. Consequently in the rath term, the coefficient of d will be n — 1. Hence, the rath term of the series, or the last term, as the number of the terms is n, will be l = a+ (n-l)d (1) 369. Given the first term, a, the last term, 1, and the num- ber of terms, n, to find the sum of the series, S. S=a+ (a + d) + (a + 2d) + + (l-2d) + (l — d) + l Writing the terms of the second member in the reverse order, S=l+ (l-d) + (I- 2 d) + + {a + 2 d) + (a + d) + a ARITHMETICAL PROGRESSION. 275 Adding these equations, term by term, we have 2S=(a+l) + (a + l) + (a + l)+ + (a+t)+(a+t) + (a+l) In this result, (a + I) is taken as many times as there are terms, or n times ; hence 2S=n(a+l),ovS= 7i (a+I) (2) Using the value of I given in (1), Art, 3G8, this may he written S=%[2a + (n-l)d] 370. 1. In the series 5, 8/11, to 18 terms, find the last term and the sum of the series. Here a = 5, n = 18 ; the common difference is always found by subtracting the first term from the second; hence d = 8-5 = 3. Substituting these values in (1) and (2), we have I = 5 + (18 - 1) 3 = 5 + 17 x 3 = 5 + 51 = 56. 1 S S = -^ (5 + 56) = 9 x 61 = 549. 2. In the series 2,-1,-4, to 27 terms, find the last term and the sum of the series. Here a = 2, n = 27, d = the second term minus the first = — 1 — 2 = — 3. Substituting these values in (1) and (2), we have 1 = 2 + (27-1) (-3) = 2 + 26 (-3) =2- 78 = -76. S=^-(2-76)=^-(-74) = 27(-37) = -999. « EXAMPLES. Find the last term and the sum of the series in the fol- lowing : 276 ALGEBRA. 3. 1, 6, 11, to 15 terms. 4. 7, 3, — 1, to 20 terms. 5. — 9, — 6,-3, to 23 terms. 6. -5,-10,-15, to 29 terms. ^ / o o . o> 7> k> to 16 terms. 8. ■=, T-p , to 19 terms. 5 15 1 5 9. - , — , to 22 terms. 2 1 10. — - , - , to 14 terms. o o 5 11. — 3, — -, to 17 terms. 113 12. j j 9 j j 3 to 35 terms. 371. Formulae (1) and (2) constitute two independent equations, together containing all the five elements of an arithmetical progression ; hence, when any three of the five elements are given, we may readily deduce the values of the other two, as by substituting the three known values we shall have two equations with only two unknown quantities, which may be solved by methods previously given. 1. The first term of an arithmetical progression is 3, the number of terms 20, and the sum of the terms 440. Find the last term and the common difference. Here a = 3, n = 20, £=440; substituting in (1) and (2), we have 1 = 3 + 19 d - 440 = 10 (3 + I), or 44 = 3 + I From the second equation, I = 41 ; substitute in the first, 41 = 3 + 19 d ; 19 d = 38 ; d = 2. ARITHMETICAL PROGRESSION. 277 2. Given d = — 3, I = — 39, S = — 264 ; find a and n. Substituting the given quantities in (1) and (2), - 39 = a + (n - 1) (- 3), or 3 n - a = 42 - 264 = - (a - 39), or a n — 39 n = - 528 From the first of these equations, a — 3 ?i — 42 ; substitute in the second, (3 n - 42) n - 39 » = - 528, or ?r - 27 n = - 176 Whence, n = - ~ = ^—^ — = 16 or 11 Substituting in the equation a = 3n — 42, When ii — 16, a = 6 n = 11, a. = — 9, Ans. The signification of the two answers is as follows : If n = 16, and a = 6, the series will be 6, 3, 0, - 3, - 6, - 9, - 12, - 15, - 18, - 21, - 24, - 27, - 30, - 33, - 36, - 39. If n = 11, and a = — 9, the series will be _ o, - 12, - 15, - 18, - 21, - 24, - 27, - 30, - 33, - 36, - 39. In either of which the last term is — 39 and the sum — 264. 113 3. Given a = ^ , d — — T x, S = — ^; find I and n. Substituting the given quantities in (1) and (2), we have J = |+(»_l)(-l),or 12l + n*=5 S_nfl '2~2 (= + l) , ox n + 3 1 n = — 9 278 ALGEBRA. From, the first of these, n = 5 — 12 I ; substitute in the second, 5 - 12 1 + 3 I (5 - 12 I) = - 9, or 36 V" - 3 1 = 14 , 3 ±y/ 9 + 2016 3 + 45 2 7 Whence, J = ^ = 70 = 3 ° r ~~ 12 Substituting in the equation re = 5 — 12 £, 9 "When I = o , n — — 3 o 1 = — — , n = 12, Ans. The first answer is inapplicable, as a negative number of terms has no meaning. Hence the only answer is, 7 l = — j2> ft = 12. Note. A negative or fractional value of n is always inapplicable, and should be neglected, together with all other values dependent on it. EXAMPLES. 4. Given d = 4, 1 = 75, re = 19 ; find a and >S f . 165 5. Given cZ = — 1, re = 15, $ = — - ; find a and Z. -j 2 6. Given a = — -, re = 18, £ = 5 ; find d and & o 3 7. Given a = — — , re = 7, $ = — 7 ; find d and Z. 8. Given I = — 31, re = 13, S= - 169 ; find a and tf. 9. Given a = 3, I = 42-f , d = 2£ ; find re and S. 10. Given a = 7, l = — 73, #=— 363 ; find re and c?. n n- 15 5 2625 11. Given a = — , d = 7 :, o= ; find re and Z. £ Ji Ji ARITHMETICAL PROGRESSION. 279 12. Given I = - 47, d = - 1, 8= - 1118 ; find a and n. 13. Given d = — 3, S — — 328, a = 2; find Z and n. 372. 2b insert any number of arithmetical means between, two given terms. 1. Insert 5 arithmetical means between 3 and — 5. This may he performed in the same manner as the examples in the previous article ; we have given the first term a = 3 ; the last term I = — 5 ; the number of terms n = 7 ; as there are 5 means and two extremes, or in all 7 terms. Substituting in (1), Art. 368, we have 4 — 5 = 3 + 6 d ; or, 6 d = — 8 ; whence, d = — -= . o 4 Hence the terms are obtained by subtracting - from 3 for 4 the first, - from that result for the second, and so on ; or, 3 5 1 1 -l - 11 -5 Ans «5j o i o j — L t — o j o" > °> sins. EXAMPLES. 2. Insert 5 arithmetical means between 2 and 4. 3. Insert 7 arithmetical means between 3 and — 1. 4. Insert 4 arithmetical means between — 1 and — 6. 5. Insert 6 arithmetical means between — 8 and — 4. 6. Insert 4 arithmetical means between — 2 and 6. 7. Insert m arithmetical means between a and b. PROBLEMS. 373. 1. The 6th term of an arithmetical progression is 19, and the 14th term is 67. Find the first term. By Art. 368, the 6th term is a + 5 d, and the 14th term is a + 13 d. Hence, 280 ALGEBRA. a+ 5d = 19 a ±13 d = 67 Whence, 8 d = 48, or d = 6 Therefore, a = — 11, Ans. 2. Find four quantities in arithmetical progression, such that the product of the extremes shall he 45, and the product of the means 77. Let a, a + d, a + 2 d, and a ± 3 d he the quantities. Then, by the conditions, a 2 ± 3 a d = 45 (1) a 2 ± 3 a d + 2 d 2 = 77 (2) Subtracting (1) from (2), 2 d 2 = 32 d 2 = 16 t?=±4. If d = 4, substituting in (1), we have a 2 + 12a = 45 _ -12±V^Ti4TT80 -12 ±18 Q Whence, a = —^ = ^ = 3 or — 15. This indicates two answers, 3, 7, 11, and 15, or, — 15, — 11, — 7, and — 3. If d = — 4, substituting in (1), we have a 2 — 12 a- 45 W1 12±V144 + 180 12 ±18 1K Q Whence, a = ! — ^ = „ = 15 or — 3. This also indicates two answers, 15, 11, 7, and 3, or, — 3, — 7, — 11, and — 15. But these two answers are the same as those obtained with the other value of d. Hence, the two answers to the problem are 3, 7, 11, and 15, or, — 3, — 7, — 11, and — 15. ARITHMETICAL PROGRESSION. 281 3. Find the sum of the odd numbers from 1 to 100. 4. A debt can be discharged in a year by paying $ 1 the first week, $ 3 the second, $ 5 the third, and so on. .Required the last payment, and the amount of the debt. 5. A person saves $270 the first year, $210 the second, and so on. In how many years will a person who saves every year $ 180 have saved as much as he ? 6. Two persons start together. One travels ten leagues a day, the other eight leagues the first day, which he augments daily by half a league. After how many days, and at what distance from the point of departure, will they come together ? 7. Find four numbers in arithmetical progression, such that the sum of the first and third shall be 22, and the sum of the second and fourth 36. 8. The 7th term of an arithmetical progression is 27 ; and the 13th term is 51. Find the first term. 9. A gentleman set out from Boston to NeW York. He travelled 25 miles the first day, 20 miles the second day, each day travelling 5 miles less than on the preceding. How far was he from Boston at the end of the eleventh day ? 10. If a man travel 20 miles the first day, 15 miles the sec- ond, and so continue to travel 5 miles less each day, how far will he have advanced on his journey at the end of the 8th day? 11. The sum of the squares of the extremes of four quanti- ties in arithmetical progression is 200, and the sum of the squares of the means is 13G. What are the quantities ? 12. After A had travelled for 2| hours, at the rate of 4 miles an hour, B set out to overtake him, and went 4£ miles the first hour, 4J the second, 5 the third, and so on, increasing his speed a quarter of a mile every hour. In how many' hours would he overtake A ? 282 ALGEBRA. XXXIV. — GEOMETRICAL PROGRESSION. 374. A Geometrical Progression is a series in which each term is derived from the preceding term by multiplying by a constant quantity, called the ratio. 375. When the series is increasing, as, for exanrple. 2, 6, 18, 54, 162, each term is derived from the preceding term by multiplying by a quantity greater than 1 ; consequently the ratio is a quantity greater than 1. When the series is decreasing, as, for example, 9 3 1 i i jl each term is derived from the preceding term by multiplying by a quantity less than 1 ; consequently the ratio is a quantity less than 1. Negative values of the ratio are admissible ; for example, -3, 6,-12,24,-48, is a progression in which the ratio is — 2. 376. Given the first term, a, the ratio, r, and the number of terms, n, to find the last term, I. The progression will be a, ar, ar 2 , ar 5 , We observe that the terms differ only in the exponent of r, which is 1 in the second term, 2 in the third term, 3 in the fourth term, etc. Consequently in the nth. or last term, the exponent of r will be n — 1, or l=ar n - x (1) 377. Given the first term, a, the last term, I, and the ratio. r, to find the sum of the series, S. GEOMETRICAL PROGRESSION. 283 S= a + a r + a r 2 + a r 3 + + a r n ~ 3 + a r"- 2 + a r n ~ x Multiplying each term by r, r S—ar+ ar 2 +ar 3 + a7 A + + a r n ~ 2 + a r" _1 + a r n Subtracting the first equation from the second, we have ft 2 ,Tl (f r S— S=ar n — a, or S (r —l) — a r n — a, or S = - — ^— But from (1), Art. 376, by multiplying each term by r, Substituting this value of a r n in the value of S, we have r — 1 378. 1. In the series 2, 4, 8, to 11 terms, find the last term and the sum of the series. Here a = 2, n = ll; the ratio is always found by dividing 4 the second term by the first ; hence, r = - = 2. Substituting these values in (1) and (2), we have I = 2 (2) 11 - 1 = 2 x 2 10 = 2 x 1024 = 2048. s= (2x204S)-2 =4096 _ 2 = 4094 Zi — JL 2. In the series 3, 1. », to 7 terms, find the last term o and the sum of the series. Here a = 3, n = 7, r = second term divided by first term = ^. Substituting these values in (1) and (2), we have l - 6 \3) ~* \3) _ 3 6_ 3 5_ 243" 284 ALGEBEA. a j_v , r , 2186 v3 X 243/ ° _729 " 729 _ 2186 3 _ 1093 ^ = ~^[ ~ - 1 — " ^2~ : T29" X 2 ~" ^43" ' 3 _1 3 ~3 3. In the series — 2, 6, — 18, to 8 terms, find the last term and the sum of the series. Here a — — 2, n = S, r= — ~ = — 3. Hence, I = (_ 2) (- 3) 8 - 1 = (- 2) (- 3) 7 = (- 2) (- 2187) = 4374. (_ 3 x 4374) - (- 2) - 13122 + 2 _^ - 13120 = (-3) — 1- -4 -4 EXAMPLES. Find the last term and the sum of the series in the fol- lowing : 4. 1, 2, 4, to 12 terms. 4 5. 3, 2, - , to 7 terms. o 6. —2, 8, —32, to 6 terms. 7. 2, — 1, ■= , to 10 terms. 111 2' V 8 8. ^ , T , q , to 11 terms. 2 3 9. k i — 1? ^ i to 8 terms. 10. 8, 4, 2, to 9 terms. 11. 4' -4' J2' to6 termS * 2 11 12. -k>— g>— gj to 10 terms. 13. 3, -6, 12, to 7 terms. GEOMETRICAL PROGRESSION. 285 379. Formulae (1) and (2) together contain all the five ele- ments of a geometrical progression ; hence, if any three of the five are given, we may find the other two, exactly as in arith- metical progression. But in certain cases the operation in- volves the solution of an equation of a higher degree than the second, for which rules have not heen given ; and in other cases the unknown quantity appears as an exponent, the solu- tion of which equations can usually only be effected by the use of logarithms ; although in certain simple cases they may be solved by inspection. 1. Given I = 6561, r — 3, n = 9 ; find a and S. Substituting these values in (1) and (2), Arts. 376 and 377, we have 6561 = a (3) 8 ; or 6561 = 6561 a ; or a = 1. (3x6561)-l_ 19683-1 _ 19682 _ 6 ~ 3=1 ~" ~2~ "" ~2~ 2. Given a = — 2, n = 5, I = — 32 ; find r and S. Substituting these values in (1) and (2), we have -32 = (-2)(r) 5 - 1 ; or-32 = -2r 1 ; r 4 = 16; r = ±2. If r = 2, S= (2 X ~ 32) 7 ( ~ 2) =- 64 + 2 = -62. T , 9 q (-2x-32)-(-2) _ 64 + 2 _66_ The signification of the two answers is as follows : If r=2, the series will be -2, -4, -8, -16, -32, in which the sum is — 62. If r = — 2, the series will be — 2, 4, — 8, 16, — 32, in which the sum is — 22. 3. Given a = 3, r = — ^ , S = ; find n and I. 286 ALGEBRA. Substituting these values in (1) and (2), we have -Us 1640 _ 3 1+9 • 6560 1 "729— ^TV=^r ; oW+9 = T29 ; ° Tl = -729- 3 3 Substituting this value of I in the equation (— 3)" _1 = -, we 3 have (— 3) n ~ 1 = ^- = — 2187; whence, by inspection, n— 1 71".) = 7, or n = 8. EXAMPLES. 4. Given J = - 256, r = — 2, n = 10 ; find a and A 5. Given r = -, n = 8, S— -^-^r ; find a and Z. o 6561 6. Given a = 2, n = 7, I = 1458 ; find r and 5. 3 7. Given a = 3, « = 6, Z = — . ,^ , ; find r and & 1024 8. Given a-= 1, r = 3, Z = 81 ; find -« and S. 1 127 9. Given a = 2, Z = ^ , $ = -^- ; find n and r. 10. Given «. = - , r = — 3, $= — 91; find n and Z. 11. Given £ = -128, r = 2, £ = -255; find n and a. 380. The Limit to which the sum of the terms of a decreas- ing geometrical progression approaches, as the number of terms becomes larger and larger, is called the sum of the scries to infinity. We may write the value of S obtained in Art. .'177 as follows : GEOMETRICAL PROGRESSION. 287 a — r I S-. 1-r In a decreasing geometrical progression, the larger the num- ber of terms taken the smaller will be the value of the last term ; hence, by taking terms enough, the last term may be made as small as we please. Then (Art. 207), the limiting value of I is 0. Consequently the limit to which the value of S approaches, as the number of terms becomes larger and larger, is - . Therefore the sum of a decreasing geometrical progression to infinity is given by the formula 1. Find the sum of the series 3, 1, ^ , to infinity. o Here a = 3, r = k ; substituting in (3), we have O 3 9 9 A 3 8 16 2. Find the sum of the series 4, — ^ , — , to infinity. _8 ~~ 3 2 Here a = 4, r = —r- = — » ; substituting in (3), we have + 3 EXAMPLES. Find the sum of the following to infinity : 3l 2 ' lj 2' 5 - - 1 '3'~9' 288 ALGEBRA. 3 11 '■ 4' 2' 3' '■" 9 8 2 1 9 - *'5' 50' 8 2 -A li in -4 4 _A 5' 35' 245' ' 5' " 25' 381. To find the value of a rejoeating decimal. This is a case of finding the sum of a geometrical progres- sion to infinity, and may be solved by the formula of Art. 380. 1. Find the value of .363636 .363636 = .36 + .0036 + .000036 + AAO/> Here a = .36, and r = '—— — = .01 ; substituting in (3), .oo .36 _ : 36_36_ £ 2. Find the value of .285151 .285151 = .28 + .0051 + .000051 + To find the sum of all the terms except the first, we have a = .0051, r = .01 ; substituting in (3), a .0051 .0051 51 17 o = 1-.01 .99 9900 3300 Adding the first term to this, the value of the given decimal 28 17 941 ~ 100 + 3300" 3300' EXAMPLES. Find the values of the following : 3. .074074 5. .7333 7. .113333. 4. .481481 6. .52121 8. .215454. GEOMETRICAL PROGRESSION. 289 382. To insert any number of geometrical means between two given terms. 64 1. Insert 4 geometrical means between 2 and p-r^ . This may be performed in the same manner as the examples 64 in Art. 379. We have a = 2, I = —-^ , and n = 6, or two more 243 than the number of means. Substituting these values in (1), Art. 376, we have 64 32 2 243 = 2? ' 5; ° rr5 = 243 ; mr = t 2 Hence the terms are obtained by multiplying 2 by ^ for the 2 first, that result by ^ for the second, and so on ; or, o 4 8 16 32 ^4 2 '3' 9' 27' 81' 243' 2. Insert 5 geometrical means between — 2 and — 128. Here a = — 2, I = — 128, n = 7. Substituting in (1), Art. 376, we have — 128 = — 2 7* ; or r 6 = 64 ; whence, r = ± 2. If r = 2, the series will be _ 2; -4, -8, -16, -32, -64, -128. If r = — 2, the series will be - 2, 4, - 8, 16, - 32, 64, - 128. EXAMPLES. o 1 128 3. Insert 6 geometrical means between o and ^Hq • 4. Insert 5 geometrical means between ^ and 364£. 5. Insert 6 geometrical means between — 2 and — 4374. 200 ALGEBRA. 729 6. Insert 4 geometrical means between 3 and — 3 1024 7. Insert 7 geometrical means between - and PROBLEMS. 383. 1. What is the first term of a geometrical progression, when the 5th term is 48, and the 8th term is — 384 ? By Art. 376, the 5th term is a r 4 , and the 8th term is a r~. Hence, ar* = 48, and a r' = — 384. Dividing the second of these equations by the first, r 3 = — 8 ; whence, r = — 2. Tl 48 48 <* J Ihen, « = — ; - = :r7r= z: 3 , Ans. r* 16 2. Find three numbers in geometrical progression, such that their sum shall be 14, and the sum of their squares 84. Let "a, a r, and a r~ denote the numbers. Then, by the con- ditions, a + a r + a r 2 = 14 (1) a 2 + a 2 r 2 + a 2 r i = 8A (2) Dividing (2) by (1), a — ar+ar 2 = 6 (3) Adding (1) and (3), a + a r 2 = 10 (4) 4 Subtracting (3) from (1), a r = 4, ofr = - (5) 1 c Substituting from (5) in (4), a -\ = 10 <r -10 a = -16 wi (K onox lOiy/100-64 10 ±6 Q _ VV hence (Art. 309), a — ~ - = — - — — 8 or 2. — Z 4 1 If a = 8, r = q = jr, and the numbers are 8, 4, and 2. 8 Z HARMONICAL PROGRESSION. 291 4 If a = 2, ?• = - = 2, and the numbers are 2, 4, and 8. Therefore, the numbers are 2, 4, and 8, Ans. 3. A person who saved every year half as much again as he saved the previous year, had in seven years saved $2059. How much did he save the first year ? 4. A gentleman boarded 9 days, paying 3 cents for the first day, 9 cents for the second day, 27 cents for the third day, and so on. Required the cost. 5. Suppose the elastic power of a ball that falls from a height of 100 feet, to be such as to cause it to rise 0.9375 of the height from which it fell, and to continue in this way diminishing the height to which it will rise, in geometrical progression, till it comes to rest. How far will it have moved ? 6. The sum of the first and second of four quantities in geometrical progression is 15, and the sum of the third and fourth is 60. Required the quantities. 7. The fifth term of a geometrical progression is — 324, and the 9th term is — 26244. What is the first term ? 8. The third term of a geometrical progression is ^-r , and 9 the sixth term is -^r^ . What is the second term ? XXXV.— HARMONICAL PROGRESSION. 384. Quantities are said to be in Harmonical Progression when their reciprocals form an arithmetical progression. For example, 1, -, F , -, 3' 5' 7 are in harmonical progression, because their reciprocals, 1, 3, 5, 7, form an arithmetical progression. 292 ALGEBRA. 385. From the preceding it follows that all problems in harmonical progression, which are susceptible of solution, may be solved by inverting the terms and applying the rules of the arithmetical progression. There will be found, however, no general expression for the sum of a harmonical series. 386. To find the last term of a given ha rmonical series. 2 2 3' 5 ; 1. In the series 2, - , p, to 36 terms, find the last term. Inverting the series, we have the arithmetical progression 2 j 2 ' 2 ' t0 36 terms • Here a = -, d = l, w = 36 ; hence, by (1), Art. 368, Z = i+(36-l)l = ^ + 35 = ^. 2 Inverting this, we obtain -=r as the last term of the given series. EXAMPLES. Find the last terms of the following : K Q A *} 1 9 2. - , - , to 23 terms. 4. r,T,-, to 26 terms. 3. - ---r, to 17 terms. 5. a, b, to n terms. 2 3 o 387. To ii/si'rf any number of harmonical means between two given terms. 1. Insert 5 harmonical means between 2 and — 3. Inverting, we have to insert 5 arithmetical means between 1 A 1 2 and ~3' HARMONICAL PROGRESSION. 293 Here a = - , Z = — -,n = J; substituting in (1), Art, 368, Z o we have — - — - + 6d: or 6 d — — -x : whence, cZ = — — . 3 2 6 ob Hence, the arithmetical means are 13 2 1_ _1_ _L 36' 9' 12' 18' 36* Then, the harmonical means will be 36 9 ._ 1Q 36 -,-,12,-lS, - T ,Ans. EXAMPLES. 2 3 2. Insert 7 harmonical means between - and — . 5 10 3. Insert 3 harmonical means between — 1 and — 5. 4. Insert 6 harmonical means between 3 and — 1. 5. Insert m harmonical means between a and b. 388. If three consecutive terms of a harmonical progres- sion be taken, the first has the same ratio to the third, that the first minus the second has to the second minus the third. Let a, b, c be in harmonical progression ; then their recip- rocals - , - , and - will be in arithmetical progression. Hence^ a> o c 1_1_1_1 c b b a' Clearing of fractions, ab — ac = ac — bc or, a (b — c) = c (a — b) Dividing through by c (b — c), we have a a — b which was to be proved. 294 » ALGEBRA. 389. Let a and c be any two quantities ; b their harrnoni- cal mean. Then, by the previous theorem, - = . Clearing of fractions, ab — ac = ac — be; then, ab + bc = 2ac 2 a c or, b = . a + c 390. We may note the following results : if a and c arc a ~f~ c any two quantities, their arithmetical mean = — - — ; their geometrical mean = \a c ; and their harmonical mean = - - . a+ c a- 2 a c a + c / / — \2 . Since X — 7. — = vV a c ) > l t follows that the product a -\- c ~j of the harmonical and arithmetical means of two quantities is equal to the square of their geometrical mean. Consequently the geometrical mean must be intermediate in value between the harmonical and the arithmetical mean. But the harmonical mean is less than the arithmetical mean, be- a + c 2ac (a + e) 2 — 4 a c a 2 + 2 ac+ c 2 — £ac cause — pr = - — T —/- r = — — r 2 a + c 2 (a + c) 2 (a + c) a 2 — 2ac + c 2 (a — c) 2 = — 7T7 ^ = 7T^ n a positive quantity. 2 (a + c) 2 (a + c) ' L >■ * Hence of the three quantities, the arithmetical mean is the greatest, the geometrical mean next, and the harmonical mean the least. XXXVI. — PERMUTATIONS AND COMBINA- TIONS. 391. The different orders in which quantities can be ar- ranged are called their Permutations. Thus, the permutations of the quantities a, b, <; taken two at a time, are PERMUTATIONS AND COMBINATIONS. 295 a b, b a ; a c, c a; be, c b; and taken three at a time, are a b c, ac b; b a e, b c a ; cab, cb a. 392. The Combinations of quantities are the different col- lections that can he formed out of them, without regard to the order in which they are placed. Thus, the combinations of the quantities a, b, c, taken two at a time, are a b, ac, be; a b, and b a, though different permutations, forming the same combination. 393. To find the number of permutations of n quantities, taken r at a time. Let P denote the number of permutations of n quantities, taken r at a time. By placing before each of these the other n — r quantities one at' a time, we shall evidently form P (n — r) permutations of the n quantities, taken r + 1 at a time. That is, the number of permutations of n quantities, taken r at a time, multiplied by n — r, gives the number of j)ermutations of the n quantities, taken r + 1 at a time. But the number of permutations of n quantities, taken one at a time, is obviously n. Hence, the number of permutations taken two at a time, is the number taken one at a time, multiplied by n — 1, or n (n— 1). The number of permuta- tions, taken three at a time, is the number taken two at a time, multiplied by n — 2, or n (n — 1) (n — 2); and so on. We observe that the last factor in the number of permutations is n, minus a number 1 less than the number of quantities taken at a tinie. Hence, the number of permutations of n quanti- ties, taken r at a time, is n(n-l) 0-2) (n-(r-l)) or, n(n — l)(n — 2) (n — r + T). (1) 296 ALGEBRA. 394. If all the quantities are taken together, r = n and Formula (1) becomes n(n-l) (n-2) 1; or, by inverting the order of the factors, 1x2x3 (n— l)n. (2) That is, The number of permutations of n quantities, taken n at a time, is equal to the product of the natural numbers from 1 up to n. For the sake of brevity, this result is often denoted by \n, read "factorial n " ; thus \n_ denotes the product of the natu- ral numbers from 1 to n inclusive. 395. To find the number of combinations of n quantities, taken r at a time. The number of permutations of n quantities, taken r at a time, is (Art. 393), n{n — l) (n — 2) (n — r+ 1). By Art. 394, each combination of r quantities produces \r permutations. Hence, the number of combinations must equal the number of permutations divided by ta or n(n-l) (n-2) (n-r+1) | r 396. The number of combinations of n quant It Its, taken r at a time is the same as the number of combinations of n quantities taken n — r at a time. For, it is evident that for every combination of r quantities which we take out of n quantities, we leave one combination of n — r quantities, which contains the remaining quantities. EXAMPLES. 397. 1. How many changes can be rung with 10 bells, taking 7 at a time ? PERMUTATIONS AND COMBINATIONS. 2! >7 Here, n = 10, r = 7 ; then n — r + 1 = 4. Then, by Formula (1), 10x9x8x7x6x5x4 = 604800, ^».s. 2. How many different combinations can be made with 5 letters out of 8 ? Here, n = 8, r = 5 ; then n — r + 1 = 4:. Then, by Formula (3), 8x7x6x5x4 1x2x3x4x5 56, Ans. 3. In how many different orders may 7 persons be seated at table ? Here n = 7 ; then, by Formula (2), 1x2x3x4x5x6x7 = 5040, Ans. 4. How many different words of 4 letters each can be made with 6 letters ? How many words of 3 letters each ? How many of 6 letters each ? How many in all possible ways ? 5. How often can 4 students change their places in a class of 8, so as not to preserve the same order ? 6. From a company of 40 soldiers, how many different pick- ets of 6 men can be taken ? 7. How many permutations can be formed of the 26 letters of the alphabet, taken 4 at a time ? 8. How many different numbers can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, taking 5 at a time, each digit occurring not more than once in any number ? 9. How many different permutations may be formed of the letters in the word since, taken all together ? 10. How many different combinations may be formed of the letters in the word forming, taken three at a time ? 298 ALGEBRA. 11. How many different combinations may be formed of 20 letters, taken 5 at a time ? 12. How many different combinations may be formed of 18 letters, taken 11 at a time ? 13. How many different committees, consisting of 7 persons each, can be formed out of a corporation of 20 persons ? 14. How many different numbers, of three different figures each, can be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, ? XXXVII. — BINOMIAL THEOREM. POSITIVE INTEGRAL EXPONENT. 398. The Binomial Theorem, discovered by Newton, is a formula, by means of which any binomial may be raised to any required power, without going through the process of invo- lution. 399. To prove the Theorem for a positive integral ex- ponent. By actual multiplication we may show that (a + x) 2 = a 2 + 2 ax + x 2 (a + x) 3 = a 3 + 3 a 2 x + 3 a x 2 + x 3 (a + x) 4 = a 4 + 4 a 3 x + 6 a 2 x 2 + 4 a x 3 + x* In these results we observe the following laws : 1. The number of terms is one more than the exponent of the binomial. 2. The exponent of a in the first term is the same as the exponent of the binomial, and decreases by one in each stic- ceedin;/ term. 3. The exponent of x in the second term is unity, and in- creases by one in each succeeding term. BINOMIAL THEOREM. 299 4. The coefficient of the first term, is unity ; and of the sec- ond term, is the exponent of the binomial. 5. If the coefficient of any term be multiplied by the expo- nent of a in that term, and the product divided by the number of the term, beginning at the left, the result will be the co- efficient of the next term,. Assuming that the laws hold for any positive integral expo- nent, n r we have . . , n(n— 1) „ „ n(n— T)(n— 2) „ , (a+x) n =a n +na n - 1 x+ \ ' a n ~ 2 x 2 + K J ±- / a n ~ s x 3 + This result is called the Binomial Theorem. 400. To prove that it holds for any positive integral expo- nent, we multiply hoth members by a + x, thus / x ^i , , n(n—T) , , n(n — l)(n— 2) . „ (a+x) n+1 =a n+1 +?ia n x + \ V ^ar^- v J \ ' a n ~ 2 x 3 , „ n(n— 1) . , + + a n x + na n ~ y x~+ v a n ~ i x 6 +. X.A = a n + 1 + (n + 1) a n x + n (n — 1) a"" 1 x- ln{n-l)(n-2) ^-1) 1 + L 1.2.3 " + 1.2 J +> = a n+1 + (w + 1) a n x + -™ 2 In - 1 + 2 1 a n ~ l a? + n S [-H <-*+ { ?i "T~ 1 ) 72- = a" +1 + (n + 1) a" a; + ^-j-tj a"- 1 ar 2 (w + 1) n (n— 1) _„ 3 1.2.3 ~ a " " * + where it is evident that every term except the first will con- tain the factor n + 1. 300 ALGEBRA. "We observe that the expansion is of the same form as the value of (a + x) n , having n + 1 in the place of n. Hence, if the laws of Art. 399 hold for any positive integral exponent, n, they also hold when that exponent is increased by 1. But we have shown them to bold for (a + cc) 4 , hence they hold for (a + x) 5 ; and since they hold for (a + x) 5 , they also hold for (a + x) 6 ; and so on. Hence they hold for any positive integral exponent. 401. Since 1.2 = [2, 1. 2. 3 = [3, etc. (Art. 394), the Bino- mial Theorem is usually written as follows : \ n(n—l) „ o n(n—l)(n—2) _ , „ (a+x) n =a n +?i a n ~ l x+ V V ~V+— £ } a n ^x 3 + If. £L 402. If a =1, then, since any power of 1 equals 1, we bave ?i(n — l) , w (n. — 1) (w — 2) „ 403. In performing examples by the aid of the Binomial Theorem, we may use the laws of Art. 399 to find the expo- nents and coefficients of the terms. 1. Expand (a + x) 6 by the Binomial Theorem. The number of terms is 7. The exponent of a in the first term is 6, and decreases by 1 in each succeeding term. The exponent of x in the second term is 1, and increases by 1 in each succeeding term. The coefficient of the first term is 1 ; of the second term, 6; if the coefficient of the second term, 6, be multiplied by 5, the exponent of a in that term, and the product, 30, be divided by the number of the term, 2, the result, 15, will be the coefficient of the third term ; etc. Eesult, a 6 +6a 5 x + 15 a 4 x 2 + 20 a 3 x* + 15 a 2 x" + 6 a x 5 + x*. BINOMIAL THEOREM. 301 Note. It will be observed that the coefficients of any two terms taken equidistant from the beginning and end of the expansion are the same. The reason for this will be obvious if, in Art. 401, x and a be inter- changed, which is equivalent to inverting the series in the second member. Thus, the coefficients of the latter half of an expansion may bo written out from the first half. 2. Expand (1 + sc) 7 by the Binomial Theorem. " Result, l 7 + 7.1 6 . x + 21.1 5 . x 2 + 35.1 4 . x 3 + 35.1 3 . x* + 21.1 s . x 5 + 7.1 1 . x° + x 1 ; or, 1 + 7 x + 21 x 2 + 35 x 3 + 35 x A + 21 x 5 + 7 x 6 + x\ Note. If the first term of the binomial is a numerical quantity, it will be found convenient, in applying the laws, to retain the exponents at first without reduction, as then the laws for coefficients may be used. The re- sult should afterwards be reduced to its simplest form. 3. Expand (2 a + 3 b)'° by the Binomial Theorem. (2a + 3&) 5 =[(2«.) + (36)] 5 = (2 a) 5 + 5 (2 a) 4 (3 b) + 10 (2 a) 3 (3 bf + 10 (2 a)' 2 (3 b) 3 + 5 (2 a) (3 by + (3 b) 5 = 32 a 5 + 240 a 4 b + 720 a 3 V 2 + 1080 a 2 b 3 + 810 a b* + 243 b 5 , Aiis. 4. Expand (irf 2 — ft -1 ) 6 by the Binomial Theorem. ( m -i _ n -i)« = [(m _i ) + (- ft- 1 )] 6 = ( m -^) +6(m"2)5(_ w -i) + i5( m -*) 4 (-«- 1 )' 2 +20(m"^) 3 (-H- 1 ) 3 + 15 (m~ty (-ft" 1 ) 4 + G (m~*) (-n- l f+ (-ft- 1 ) 6 ' =mr 3 + 6 m~ r% (- ft" 1 ) + 15 m~ 2 (ft- 2 ) + 20 i»~* (- ft" 3 ) + 15 m- 1 (ft- 4 ) + 6 m~- (- ft" 5 ) + (ft" 6 ) = m~ 3 — 6 m~* ft- 1 + 15 m~- n~- — 20 m~ - ft~ 3 + 15 m~ x ft" 4 — 6 m " 2 ft -5 + ft -6 , Ans. 302 ALGEBRA. Note. If either term of the binomial is not a single letter, with unity as its coefficient and exponent, or if either term is preceded by a minus sign, it will be found convenient to enclose the term, sign and all, in a parenthesis, when the usual laws for exponents and coefficients may be applied. In reducing, care must be taken to apply the principles of Arts. 227 and 259. EXAMPLES. Expand the following by the Binomial Theorem : 5. (1 + o) 5 . 8. (a b — c d)~. 11. (J + <$y. 6. (a + a- 3 ) 6 . 9. O 2 + 3 nr)\ 12. (m~% + 2 ?i s y. 7. (x*-2yy. 10. (a- 2 - 4 x*) b . 13. (a-i — &»aj4)«. 404. To find the rth or general term of the expansion of (a + ,r)\ "We have now to determine, from the observed laws of the expansion, three things ; the exponent of a in the term, the exponent of x in the term, and the coefficient of the term. The exponent of x in the second term is 1 ; in the third term, 2 ; etc. Hence, in the rth term it will be r — 1. In any term the sum of the exponents of a and x is n. Hence, in the rth term, the exponent of a will be such a quan- tity as when added to r — 1, the exponent of x, will produce n ; or, the exponent of a will be n — r + 1. The coefficient of the term will be a fraction, of the form n(n-l)(»-2) , . , , . , , -. — - } — q— ; m which we must determine the last X- . w . o factors of the numerator and denominator. We observe that the last factor of the numerator of any tennis 1 more than the exponent of a in that term: hence the last factor of the numerator of the rth term will be n — r + 2. Also, the last factor of the denominator of any term is the same as the exponent of x in that term : hence the last factor of the denominator of the rth term will be r — 1. BINOMIAL THEOREM. 303 Therefore the n (n - 1) (n - 2) (n — r + 2) rth term = -^ t^t;-^ — ^^^rr - 1 — ct n ~ r + 1 x r ~\ 1 . Z . 6 (r — 1) 1. Find the 8th term of (3 J -2b~ l ) n . Here r = 8, n = 11 ; hence, the 8thtom = 1 l:2°3 9 4 8 5.6 6 7 5 < 3a *> , (- 2t -)' = 330 (81 a 2 ) (- 128 b~'') = - 3421440 a 2 J- 7 , ^ras. Note. The note to Ex. 4, Art. 403, applies with equal force to examples in this article. EXAMPLES. Find the 2. 10th term of (a + x) 15 . 5. 5th term of (1 - a 2 ) 12 . 3. 6th term of (1 + m) u . 6. 9th term of (x~ 1 -2 y*) u . 4. 8th term of (c - d) l \ 7. 8th term of (a% + 3 x" 1 ) 10 . 405. A trinomial may he raised to any power by the Bi- nomial Theorem, if two of its terms he enclosed in a paren- thesis and regarded as a single term ; the operations indicated being performed after the expansion by the Theorem has been effected. 1. Expand (2 a — b + c 2 ) 3 by the Binomial Theorem. (2 a - b + c 2 ) 3 = [(2 a-b) + (c 2 )] 3 = (2 a - by + 3 (2 a -b) 2 (c 2 ) + 3 (2 a - b) (c 2 ) 2 + (c 2 ) 3 =Sa 3 ~12a 2 b + 6ab 2 -b 3 +3c 2 (4:a 2 -4ab + b 2 ) + 3c 4 (2a-b) + c 6 = 8 a 3 - 12 a 2 b + 6 a b 2 - b 3 + 12 a 2 c 2 - 12 a b c 2 + 3 b 2 c 2 + 6ac i -3bc 4 + c 6 , Ans. 304 ALGEBRA. The same method will apply to the expansion of any poly- nomial hy the Binomial Theorem. EXAMPLES. Expand the following by the Binomial Theorem : 2. (1-x-x 2 )*. 4. (1 - 2 x - 2 x 2 )\ 3. (x-+3x + l) 3 . 5. (l + a;-a; 2 ) 5 . XXXVIII.— UNDETERMINED COEFFICIENTS. 406. A Series is a succession of terms, so related that each may be derived from one or more of the others, in accordance with some fixed law. The simpler forms of series have already been exhibited in the progressions. 407. A Finite Series is one having a finite number of terms. 408. An Infinite Series is one whose number of terms is unlimited. The progressions, in general, are examples of finite series ; but, in Art. 380, we considered infinite Geometrical series. 409. An infinite series is said to be convergent when the sum of the first n terms cannot numerically exceed some finite quantity, however large n may be ; and it is said to be diver- gent when the sum of the first n terms can numerically exceed any finite quantity by taking n large enough. For example, consider the infinite series 1 + x + x- + X 3 + The sum of the first n terms 1 + x + x 2 + x 3 + +x n ~ 1 = ^^ (Art. 120). 1 — x UNDETERMINED COEFFICIENTS. 305 If x is less than 1, x n is less than x, however largo n may be ; consequently the numerator and denominator of the frac- tion are each less than 1, and positive ; and the numerator is larger than the denominator ; hence the fraction is equal to some finite quantity greater than 1. The series is therefore convergent if x is less than 1. If x is equal to 1, each term of the series equals 1, conse- quently the sum of the first n terms is n ; and this can numer- ically exceed any finite quantity by taking n large enough. The series is therefore divergent if x = 1. If x is greater than 1, each term of the series after the first is greater than 1, consequently the sum of the first n terms is greater than n ; and this sum can numerically exceed any finite quantity by taking n large enough. The series is therefore divergent if x is greater than 1. 410. Every infinite literal series, arranged in order of pow- ers of some letter, is convergent for some values of that letter, and divergent for other values. We will now show that it is convergent when that letter equals zero. Let the series be a + bx + cx 2 + dx s + + Jex n ~ 1 + The sum of the first n terms is a + bx + cx 2 + dx 3 + + kx n ~\ which is equal to a, if x is made equal to 0. Hence, however large n may be, the sum of the first n terms is equal to a, if x is equal to 0. Consequently the series is convergent if x = 0. 411. Infinite series may be developed by the common pro- cesses of Division, as in Art. 101, Exs. 19 and 20, and Extrac- tion of Roots, as in Arts. 239 and 243 ; and by other methods which it will now be our object to elucidate. 306 ALGEBRA. UNDETERMINED COEFFICIENTS. 412. A method of expanding algebraic expressions into series, simple in its principles, and general in its application, is based on the following theorem, known as the THEOREM OF "UNDETERMINED COEFFICIENTS. 413. If the series A + Bx + Cx 2 + Bx 3 + is always equal to the series A' + B' x -f- C ar -f- D' x 3 + , for a ny value of x which makes both series convergent, the coefficients of like powers of ' x in the two series will he equal. For, since the equation A + Bx + Cx 2 +Dx 3 + =A'+B'x+ C'x-+D>x 3 + is satisfied for any value of x which makes both members con- vergent ; and since by Art. 410, if x is equal to 0, both mem- bers are convergent ; it follows that the equation is satisfied if x = 0. Making x — 0, the equation becomes A = A'. Subtracting A from the first member of the equation, and its equal, A', from the second member, we have Bx+ Cx 2 +Bx 3 + = B'x + C'x 2 + D'x 3 + Dividing through by x, B+Cx + Bx 2 + =B'+ C'x + D'x 2 + This equation is also satisfied for any value of X which makes both members convergent ; hence it is satisfied if x = 0. Mak- ing x = 0, we have; B = B'. Proceeding in this way, we may show C= O, I) = D', etc. UNDETERMINED COEFFICIENTS. 307 Note. The necessity for the limitation of the theorem to values of x which make both series convergent, is that a convergent series evidently cannot be equal to a divergent series ; and two divergent scries cannot be equal, as two quantities which numerically exceed any finite quantity can- not be said to be equal. Hence, in all applications of the theorem, the results are only true when both members are convergent. APPLICATION TO THE EXPANSION OF FRACTIONS INTO SERIES. 2 + 5 x 414. 1. Expand - — - into a series. JL — o x We have seen (Art. 101), that any fraction may he expanded into a series by dividing the numerator by the denominator ; consequently, we know that the proposed development is pos- sible. Assume then, ^ + f x =A+Bx + Cx 2 + Dx*+Ex i + (1) 1 — ox where A, B, C, D, E, are quantities independent of x. Clearing of fractions, and collecting together in the second member the terms containing like powers of x, we have 2+5x = A + B -3.4 x+ C -3B x-+ D -3(7 x 3 + E -3D x* + Equation (1), and also the preceding equation, are evidently to be satisfied by all values of x which make the second mem- ber a convergent series. Hence, applying the Theorem of Un- determined Coefficients to the latter, we have A = 2. B — 3 A — 5 ; whence, B = 3 A + 5 = 11. C-3B = 0; whence, C = 3B =33. D - 3 C = ; whence, D = 3 C = 99. E-3D = 0; whence, E=3D =297. 303 ALGEBRA. Substituting these values of A, B, C, D, E, in (1), we have ^+-^ = 2 + 11 a: + 33 z 2 + 99 cc 3 + 297 a: 4 + , 1 — o x which may he readily verified by division. This result, in accordance with the last part of the Note to Art. 413, only expresses the value of the fraction for such values of x as make the second member a convergent series. 2. Expand — — « into a series. l-2ic-a;' 2 Assume 1 — 3 x- ■x* A + Bx + Cx 2 +Dx !i + JEx i + l—2x—x Clearing of fractions, and collecting terms, 1-Sx x 2 = A + B -2 A x+ C -2B - A x 2 + D -2(7 - B x 3 + E -2D - C .v Equating the coefficients of like powers of x, A = l. B-2A = -3; whence, B — 2A — 3 = — 1. C-2B-A = -1; whence, C=2B + A-l = -2. D-2C—B= 0; whence, D=2 C+B = — 5. E-2D- 0— 0; whence, E=2D+ C=-12. Substituting these values, 1-3 x ■x \-2x ■X" r = 1 — x — 2 x 2 — 5 x 3 — 12 x* — , A ns. Note. This method enables us to find the law of the coefficients in any expansion. For instance, in Example 1, we obtained the equations C=3B, D = BC, E = ZD, etc. ; or, in general, any coefficient, after the second, is three times the preceding. In Example 2, we obtained the equations Ij — IC-^B, E=2D + C, etc.; or, in general, any coefficient, after the third, is twice the preceding plus the next but one preceding. After the law of the coefficients of any expansion has been found, we may write out the subsequent terms to any desired extent by its aid. 2-3 x + Ax 2 1 + 2 x — 5x 2 ' 3 + x -2x 2 3 — x + x 2 ' 1- -3x 2 UNDETERMINED COEFFICIENTS. 309 EXAMPLES. Expand the following to five terms : 3. * = ". 6. \~'-* . 9. 1 + X 1 + x + x z 3+1* 7 l-^ 2 1Q l-5x' l + 2x-3a; 2 ' 2 - a; + x 2 8 l + 2.x n Sl 1-x 2 ' ' 2-cc-a; 2 ' " 2 -3 a: -2 a- 415. If the lowest power of x in the denominator is higher than the lowest power of x in the numerator, the method of the preceding article will he found inapplicable. We may, how- ever, determine by actual division what will be the exponent of x in the first term of the expansion, and assume the fraction equal to a series whose first term contains that power of x ; the exponents afterwards increasing by unity each term as usual. 1. Expand ^ ; 2 into a series. O X — X Proceeding in the usual way, we should assume 1 =A+Bx+ Clearing of fractions, l = 3Ax + (3B — A)x 2 + Equating the coefficients of like powers of x, we have 1 = ; a result manifestly absurd, and showing that the usual method is inapplicable. x- 1 But, dividing 1 by 3 x — x' 1 , we obtain — — as the first term o of the quotient ; hence we assume the fraction equal to a series whose first term contains x~ l ; next term x°, or 1 ; next term x ; etc. Or, 5 T = Ax~ 1 + B + Cx + Dx 2 + Ex*+ OX — X' 310 ALGEBRA. Clearing of fractions, and collecting terms, x* + 1=3A+3B - A x + 3C - B x- + oD - C x 3 + 3E - D Equating the coefficients of like powers of x, 3 A = 1 ; whence, A = - . o 3B~A = 0- whence, B = — = - . 3 9 7? 1 3 C—B — 0] whence, C = — = — . 3D- C=0; whence, D = ^- = — . 3E-D = 0; whence, E=^ = ~. Substituting these values, 1 1.11 1.1, 3x-x 2 3 9 27 '81 '243 EXAMPLES. Expand the following to five terms : 2 n l + z-a 2 . l — 2x* — at 3sc 2 -2a; 8 " cc-2a 2 +3cc 3 a- a + a: 8 -a- 4 APPLICATION TO THE EXPANSION OF RADICALS INTO SERIES. 416. As any root of any expression consisting of two or more terms can be obtained by the method of Art. 247, we know that the development is possible. 1. Expand \'l + x 2 into a series by the Theorem of Unde- termined Coefficients. Assume \Jl-\-x' 2 = A + Bx + Cx*+ Dx* + 22x i + UNDETERMINED COEFFICIENTS. 311 Squaring both members, we have (Art. 230), l + x 2 = A 2 + 2AB x + B 2 + 2 AC .'■- + 2 AD + 2BC x 3 + C 2 + 2AE + 2BD ./■• Equating the coefficients of like powers of x, A 2 = l; whence, A = l. 2AB = 0; whence, B = - B 2 + 2AC=1; whence, C = 2 A 1 - B 2 _ 1 ^2 2vlZ> + 2i?C Y = 0; whence, D = =0 C 2 + 2AB+2BD = 0; whence, .E=- 2 A BC A 2BD+ C 2 2 A 1 8 Substituting these values, yi + x 2 = l + -x 2 --;x i + 2 8 which may be verified by the method of Art. 239. Note. From the equation A- = \, we may have A= ± 1 ; and taking -4 = - 1, we should find C = - — , E=— , , so that the expansion might 2 8 he as follows : 2 8 This agrees with the remark made after the rule in Art. 239. EXAMPLES. Expand the following to five terms : 2. Sjl + x. 4. v / l-2a; + 3x 2 . 3. v/1-2*. 5. y/1 + * + x 2 . 6. yi-aj. 7. yl + ic + a; 2 . [12 ALGEBRA. APPLICATION TO THE DECOMPOSITION OF RATIONAL FRACTIONS. 417. When the denominator of a fraction can be resolved into factors, and the numerator is of a lower degree than the denominator, the Theorem of Undetermined Coefficients ena- bles us to express the given fraction as the sum of two or more 'partial fractions, whose denominators are the factors of the given denominator. We shall consider only those cases in which the factors of the denominator are all of the first degree. CASE I. 418. When the factors of the denominator are all un- equal. x + 7 . Let — tt— 7z ^r be a fraction, whose denominator is (3 x — 1) (5 x + 2) composed of two unequal first degree factors. We wish to prove that it can be decomposed into two fractions, whose denominators are 3 x — 1 and 5 x + 2, and whose numerators are independent of x. To prove this, assume x + T A B + (3 x - 1) (5 x + 2) 3 x - 1 5 x + 2 We will now show that such values, independent of x, may be given to A and B, as will make the above equation identi- cal, or true for all values of x. Clearing of fractions, x + 7 = A (5 x + 2) + B (3 x - 1) or, x + 7 = (5 A + 3 B) x + 2 A - B, which is to be true for all values of x. Then, by Art. 413, the coefficients of like powers of x in the two members must be equal ; or, 5^+35=1 2A-B=7 UNDETERMINED COEFFICIENTS. 313 From these two equations we obtain A = 2, and B = — 3. Hence, the proposed decomposition is possible, and we have x + 7 2 -3 (3x-l)(5x + 2) 3x-l'5x + 2 2 o 3a;-l 5x + 2 This result may be readily verified by finding the sum of the fractions. In a similar manner we can prove that any fraction, whose denominator is composed of unequal first degree factors, can be decomposed into as many fractions as there are factors, having these factors for their denominators, and for their nu- merators quantities independent of x. EXAMPLES. 1. Decompose — 2 — j- — — — into its partial fractions. The factors of the denominator are x — 8 and x — 5 (Art. 118). a .i 3 x — 5 A B ,* >. Assume, then, — — — = -\ (J-) x- — Id x + 40 x — 8 x — 5 Clearing of fractions, and uniting terms, 3 x - 5 = A (x - 5) + B (x - 8) 19 Putting x = 8, 19 = 3 A, or A = -g-- Putting cc = 5, 10= — 3 B, or B= -— g-- Note. The student may compare the above method of finding A and B with that used on page 312. Substituting these values in (1), 19 10 3^-5 "3" 3_19 10 x 1 - 13 x + 40 ~~ x-S + x — 5 ~ 3 (x - 8) ~ 3 (x - 5) ' AnS ' 314 ALGEBRA. EXAMPLES. Decompose the following into their partial fractions 5ic — 2 3a; + 2 n cc «• - o — r • ^ i — o — • ar — 4 ar — 2 a; x + 9 _ 2a; — 3 0. 6. •> ar — 13 a- + 42 ' 7. 17 6 a; 2 - 13 x - 5 a; * x 1 + 3 a; ' a; 2 — 3 a; — 4 ft 7a ' + 9 Q 9 + 9 x - 4 a; 2 * * (a; 2 - 1) (x-2)' CASE II. 419. When the factors of the denominator are all equal. x 2 — 11 x + 26 1. Separate '- 3 — into its partial fractions. \x — o) If we attempt to perform the example hy the method of Case I, we should assume x*—l±x + 2& ABC + o + (x — 3) 3 x — 3 x — 3 x — 3 This would evidently he impossible, as the sum of the frac- tions in the second member is — ; which, as A, B, x — o and C are, by supposition, independent of x, cannot be equal a; 2 -11 a; + 26 tO ; jr-r . (x — 3) 8 The method to be used in Case II depends on the following : n ., .,. a x"- 1 + b x n ~ 2 + c x"-* + + k Consider the traction r . (aj + h) n Putting x = y — h, the fraction becomes a(y- hy- 1 + b(i/ — h) n - 2 + e (y — A)"" 3 + + k UNDETERMINED COEFFICIENTS. 315 If the terms of the numerator are expanded by the binomial theorem, and the terms containing like powers of y collected together, we shall have a fraction of the form a, y' 1 ' 1 + b r y n ~ 2 + c x y n ~ 3 + +ky y , Dividing each term of the numerator by y n , we have a x b x c x l\ y y y v Changing back y to x + h, this becomes «i b, c, /.', + , , ' , + , ,' Q + + x + h (x+ h) 2 (x + h) 3 (x + h) n This shows that the assumed fraction can be expressed as the sum of n partial fractions, whose numerators are indepen- dent of x, and whose denominators are the powers of x + h, beginning with the first, and ending with the nth. In accordance with this, we assume a 2 -11 a; + 26 A B , C ( x -3) 3 ~ x - 3 T (x - 3) 2 ^ (x - 3) 3 ' Clearing of fractions, x 2 -llx + 26 = A(x-3)- + B(x-3)+ C = A (x 2 -6x + 9) + B(x — B)+-C = A x 2 + (B - 6 A) x + 9 A - 3 B + C. Equating the coefficients of like powers of x, A = l, B-6A = -11, and 9A-3B+ C=26 Whence, A = 1, B = -5, and C = 2. Substituting these values, a; 2 -lla; + 26_ 1 5 2 (x - 3) 3 ~~ " ~x~^3 ~ (x - 3) 2 + (x - 3) 3 ' US ' 2' 316 ALGEBEA. EXAMPLES, Separate the following into their partial fractions : ar+3a: + 3 a; 2 3 a; — 10 (x + 1) 3 " ' (x-2) 3 * (2a--5) s 2 a; -13 3 a; 2 -4 18 a; 2 + 12 a; - 3 (x-5) 2 ' (cc + 1) 8 ' (3 x + 2/ CASE III. 420. When some of the factors of the denominator are equal. 1. Separate zrrz into its partial fractions. 1 x (x + l) 3 l The method in this case is a combination of the methods of Cases I and II. We assume 3 a + 2 A B C D + ,.. . ,,, + ,.. , ^„ + — • a; (x + l) 3 a- + 1 (x + l) 2 (a; + l) 3 Clearing of fractions, 3ai + 2 = A x (x + l) 2 + B x {x + 1) + Cx + D (x + l) 3 = (A + D) x* + (2A + B + 3 D) x 2 + ( A +B + C+ 3 D) x +B. Equating the coefficients of like powers of x, £> = 2, A + B + C+3 B = 3, 2A + B+3B = 0, and A + B = Whence, A = -2, B = -2, C = l, andZ> = 2. Substituting these values, 3a; + 2 2 2 12, a;(a; + l) 3 a; + 1 (a? + 1) 2 (a- + l) 3 x It is impossible to give an example to illustrate every pos- sible case; but no difficulty will be found in assuming the UNDETERMINED COEFFICIENTS. 317 proper partial fractions, if attention be given to the following general case. A fraction of the form X (x + a) (x + b) (x + m) r (x + n) s should he put equal to A B E t F K x+a x+b x + m (x+m)' (x + m) r L M R + x + n + (x + n) 2 + + (x + n) s + Single factors, like x + a and x + b, having single fractions A B like and - — — , corresponding ; and repeated factors, x+a -x+b like (x + m) r , having r partial fractions corresponding, ar- ranged as in Case II. EXAMPLES. Separate the following into their partial fractions : 8-3 x-x' 2 15 - 7 x + 3 x 2 - 3 x z x(x + 2) 2 ' x 3 (x + 5) 3 ,-r 3 - 11 x 2 + 13 x - 4 6 ar - 14 x + 6 x (x - 1) (x - 2) 2 ' (x-2)(2x-3y 2 ' 3a;-l ? 5 x 2 + 3 x + 2 x 2 {x + iy ' x 3 (x + iy 2 ' 421. Unless the numerator is of a lower degree than the denominator, the preceding methods are inapplicable. :' .'•- + 1 . For example, let it be required to separate -— ^— — into its partial fractions. Proceeding in the usual way, we assume 2 .r- +1 A B Ob *jC JC *fc -x. Clearing of fractions, 2 x 2 + 1 = A (x - 1) + B x = (A + B) x - A. 318 ALGEBRA. Equating the coefficients of like powers of x, we have 2 = 0; an absurd result, and showing that the usual method is inap- plicable. But by actual division, as in Art. 150, we have 2z 2 +l 2« + l — £ + x- — x x- X 2 x + 1 . .,„.-, We may now separate — ^ into its partial fractions by the usual method, obtaining 2a; + 1 1 3 x' — x X X 1" Hence, 2a 2 +l 2^ + 1 1 3 = 2 + —r- =2- - + - —,Ans. x- X x- — X APPLICATION TO THE REVERSION OF SERIES. 422. 1. Given y = 2x + x 2 — 2x z — 3x* + , to revert the series, or to express x in terms of y. Assume x = Ay + B if + G if + D if + (1) Substituting in this the given value of y, we have x=A(2x + x 2 -2x s -3x i +...)+B(ix 2 + x 4 +Ax s -Sx i + ...) + <7(8x 3 + 12z 4 +...) + Z>(16z 4 +...) + or, X: 2Ax+ A x 2 -2A x 3 - 3 A + 4:B + 4B - IB + 8 + 12 G + 16Z> x* + Equating the coefficients of like powers of x, 2 A = 1 ; whence, A = - . A + 4 B— ; whence, B— r = — - , 4 8 -2A + 4.B+S G=0; whence, 0= 3^ 16" -3^-7^+12 C+1GZ> = 0; whence, D- — 13 128' UNDETERMINED COEFFICIENTS. 319 Substituting these values in (1), V if 3 y 3 13 y i If the even powers of x are wanting in the given series, we may abridge the operation by assuming x equal to a series containing only the odd powers of y. Thus, to revert the series y = x — x 3 + x 5 — x 1 + , we assume x = A y + B y 3 + C y 5 + D y" + If the odd powers of x are wanting in the given series, the reversion of the series is impossible by the method previously given. But by substituting another letter, say t, for x~, we may revert the series and obtain a value of t, or of x 2 , in terms of y ; and by taking the square root of the result, express x itself in terms of y. If the first term of the series is independent of x, we cannot, by the method previously given, express x definitely in terms of y ; though we can express it in the form of a series in which y is the only unknown quantity. 2. Kevert the series y = 2 + 2x — x 2 — x 3 + 2x i + We may write the series, y-2 = 2x-x 2 -x 3 + 2x 4 + (1) Assume x=A(y-2) + B(y-2y 2 +C(y-2) 3 +D(y-2y+... (2) Substituting in this the value of y — 2 given in (1), we have x = A(2x— x 2 — x s +2x*+ ...) + B(4:X 2 + x i — ix 3 — 4x*+ ...) + C(Sx 3 -12x i + ...) + D(lGx 4 + ...)+ or, x = 2 A x — A + 4:B x 2 - A x 3 + 2 A -42? - 3B + SC -12C + 16B x* + 320 ALGEBRA. Equating the coefficients of like powers of x, 2 A — l; whence, A = ^ . — A + 4 B =:Jd ; whence, B = ^ . o — A-4:B + 8C = 0; whence, C = ^. o 7 2A-3B-12C+16I> = 0] whence, D = -j-. Suhstituting in (2), x=l(y-2)+l( 1/ -2y+l( }/ -2y+^( u -2y+ ,Am. EXAMPLES. Revert the following series to four terms : 3. y = x + x- + x 3 + x i + 4. ij = 2x + 3x 3 + 4:x 5 + 5x 7 + 5. i/ = x — x 3 + x 5 — x 1 + 2 3 4 ^y»— rj/*" 'V*^ \K/ *Aj *AJ tAs \Kj \Aj 8. y = 3x-2x 2 +3x 3 -4x i + Note. This method may sometimes be used to find, approximately, the root of an equation of higher degree than the second. Thus, to solve the equation we may put .1 = 2/, and revert the series ; giving, as in Ex. 1, Art. 422, 1 1 „ 3 , 13 x = — ii if-\ if if + 2 & J 16 128 Putting back y=.l, we have .1 .01 .003 .0013 ^ 2 8 16 128 = . 05-. 00125 + . 00019 -. 00001 + = .04893 + , Ans. This method can, of course, only be used when the series in the second member is convergent. BINOMIAL THEOREM. 321 XXXIX. — BINOMIAL THEOREM. ANY EXPONENT. 423. We have seen (Art. 402) that when n is a positive integer, n(n— 1) „ n(n — l)(n — 2) „ (l + a; )»=l + wa; + — ^r V+ — ^ '- x z + We shall now prove that this formula is true when n is a positive fraction, a negative integer, or a negative fraction. 1. Let n be a positive fraction, which we will denote by p ... — ; p and q being positive integers. Now (Art. 252), (1 + x) * = V (1 + xf = yi+px+ , (Art. 402). Extracting the ^th root of this expression by the method of Art. 247, 1 +px + 1« = 1 . p x 1 + — + q j) x That is, (l + a:)i" = l + — + (1) 2. Let n be a negative quantity, either integer or fraction, which we will denote by — s. Then (Art. 255), = , (by Arts. 402, and 423, 1). 1 + sx+ ' v J From which, by actual division, we have (l + x)-° = l-sx + (2) 322 ALGEBRA. From (1), (2), and Art. 402, we observe that whether n is positive or negative, integral or fractional, the form of the expansion is (1 + x) n = 1 + n x + A x 2 + B x 3 + Cx* + (3) x Writing - in place of x, we have a /y* /y>-J /y.3 /y»4 1 + -) =l + n- + A~ r2 + B-+C— i + al a a 1 a 6 a 4 Multiplying this through by a n , and remembering that (x\ n r / x\ i " 1+-) = a\l+-j = (a + x) n , we have (a + x) n = a n + n a"' 1 x + A a n ~ 2 x- + Ba n ~ s x s + (4) To find the values of A, B, etc., we put x + z for x in (3), and regarding (x + z) as one term, we shall have [1 + (x + z)'] r ' = l + ii(x + z) + A(x + z) 2 + B(x + z) 3 + = 1 + n x + A x' 1 + B x s + + (n + 2Ax + 3Bx 2 + )*+ (5) Regarding (1 + x) as one term, we shall have, by (4), [(1 + x) +z~] n = (1 + x) n + n (l + x) n - 1 z+ (6) Since [1 + (x + z)~\ n = [(1 + x) + z\ n , identically, we have from (5) and (6), l + nx + Ax 2 +Bx z + + (n + 2Ax + 3Bx 2 + )z+ = (1 + a-) n + w(l + a-) n_1 * + which is true for all values of z which make both members of the equation convergent. Hence, by Art. 413, the coefficients of z in the two series must be equal ; or, 11 (l + x) n - 1 = n + 2Ax + 3Bx 2 + BINOMIAL THEOREM. 323 Multiplying both members by 1 + x, n(l + x) n = n+ (2A + n)x+(3B+2A)x 2 + or, by (3), n + n 2 x + n Ax 2 + n Bx 3 + = n+ (2 A + n) x + (3B+2A)x 2 + •which is true for all values of x which make both members of the equation convergent ; hence, equating the coefficients of like powers of x, sfy {fl, j\ 2 A + n = n 2 ; whence, 2 A = n 2 — n, or A = - --= 3B+2A = ?iA; whence, 3 B=n A-2 A = A (n-2) „ A (n - 2) n (n - 1) (n - 2) B =~3— = \3 Substituting in (4), (a + x) n = a n + na"- 1 x+ V ^\~ ' a n ~ 2 x 2 n(n-l)(n-2) + — i -^ a n ~ a X s + which has tbus been proved to hold for all values of n, positive or negative, integral or fractional. Hence, the Binomial The- orem has been proved in its most general form. The result, however, only expresses the value of {a + x) n for such values of x as make the second member convergent (Art. 413). 424. When n is a positive integer, the number of terms in the expansion is n + 1 (Art. 399). When n is a fraction or negative quantity, the expansion never terminates, as no one of the quantities n — 1, n — 2, etc., can become equal to zero. The development in that case furnishes an infinite series. 324 ALGEBRA. 425. The method and notes of Art. 403 apply to the ex- pansion of expressions hy the Binomial Theorem when the exponent is a fractional or negative quantity. 2 1. Expand (a + x) :! to five terms. 2 The exponent of a in the first term of the expansion is - , and o decreases hy one in each succeeding term. The exponent of x in the second term of the expansion is 1, and increases hy one in each succeeding term. The coefficient of the first term is 1 ; of the second term, 2 2 - ; multiplying the coefficient of the second term, -, hy the ex- o o 1 2 ponent of a m that term, — - , and dividing the product, — - , hy the numher of the term, 2, we ohtain — - as the coefficient of the third term ; etc. 2 2_i 1 _4 4 -i 7 _jo Kesult, a 3 + ~a 3 x—-a ^x 2 +^ra 3 x 3 — 7 r-r^a s x A -\- .^1-2 2. Expand (1 + 2 a: 2 ) -2 to five terms. (l + 2zV 2 =D- + (2*-)]- = l- 2 - 2.1-= . (2 x$) + 3.1- 4 . (2 xfy - 4.1- 5 . (2 a;*) 8 + 5.1- 6 . (2 a;*) 4 i = 1 - 2 (2 x-) + 3 (4 a) - 4 (8 x?) + 5 (16 a 2 ) - .... = l-4a 2 + 12x-32a:- + S0a; 2 — , Ans. 3. Expand (a -1 — 3 a: *) * to five terms. (a- 1 -3a 5 ~ i )~* = [(«" 1 ) + (-3 a; - *)]"* BINOMIAL THEOREM. 325 = (a- 1 )"^ - \ (co- 1 )-'" (- 3 aT*) + " (0~ V (- 3 x~fy 1 40 L3 _ JL 455 —JUS 1 -TFr(0~ v (-3* l ) 8 +iS(«- 1 ) ¥ (-3* *)«- 81 v y v y ' 243 447 _i 14 lo 140 13 _a = ft 3_* :w 3a; 2 ) + -3 (9 a; -i)__^ a V(_27 a; 2) O 9 ol +i-*>o- 4 7 _i J-" 140 13 _a 455 .lb = a^+4a 3 a; 2 + 14a 3 cc _1 + — — a 3 x 2 + -— - a 3 #~ 2 , o o ^l?zs. EXAMPLES. Expand the following to live terms : 4. O + z) 2 8. . 8/r — — ' 12. (m~"3-_2« 2 ) -2 . v y yl + a; v ' 5. (1 + ^)- 6 . 9. ^^y 3 - 13. (l + e.r 1 )^. 6. (l-x)~%. 10. — — . 14. (a; 4 + 4aJ)t c 2 + d 7. V^ 1 ^- 11. (aT*-8y)*. 15- (g -i_ 8y -y 426. The expression for the rth term, derived in Art. 404, holds for any value of n, as it was deduced from the expansion which has been proved to hold universally. 1. Find the 7th term of (1 — cc)~ 3 . Here r = 7, n = — ^ ; hence, the o 1 4 _7 10 13 16 ~3 ,_ 3*~3'""3"' ' "3"" " 3 . NR 728a; 6 j th term = 17 2.3.4.5.6 ( ~ x) = "656T ' Ans. 326 ALGEBRA. 2. Find the 8th term of (eft + x~%)~ z . Here r = 8, n = — 3 ; hence, the -3.-4.-5.-6. — 7.-8.-9 , ^ „ . _* , 8thterm = 0.3.4.5.6.7 ^- W (* ^ = — 36 ar b x 3 , Ans. EXAMPLES. Find the 3. 8th term of \/ a + x. 7. 7th term of (x' 1 — ifift. 4. 7th term of (1 + m)- 4 . 8. 5th term of . (n~* — <ry 5. 5th term of (1 — ar)~ *. 9. 6th term of (eft + 3 x~ l )~%. 1 -2 6. 6th term of . 10. 8th term of (x 3 y — z 3 ) -3 . y'x 2 + f 427. To find any root of a number approximately by the Binomial Theorem. 1. Find the approximate square root of 10. y/ 10 = 10* = (9 + 1) * = (3 2 + 1)* Expanding this hy the Binomial Theorem, (3' + 1)* = (3*)* + \ (3 a )-i - \ (3')-* + 1 (3T* 111 ^ — 3 _L Z 3-1 _ Z Q-8 i Q-5 ^ q-7 i -6 + 2 .6 8 .3 +jg. 3 -J28-3 + = 3-i _1_ J_ _t_ _JL + 2.3 8.3 3 + 16.3 6 128. 3 7+ = 3 + .16667 - .00163 + .00026 - .00002 + = 3.16228+, BINOMIAL THEOREM. 327 which is the approximate square root of 10 to the fifth decimal place, as may be verified by evolution. 2. Find the approximate cube root of 26. $ 26 = 26* = (27 -1)3 = (3 3 - 1)* Expanding this by the Binomial Theorem, (3 3 - 1)* = (3 3 )^ + \ (3 3 )- 3 * (- 1) - i (3 3 )"* (- 1)» +| [ (3 3 r l (-i) 3 - 1 1 K Q Q-2 Q-5 Q-8 -°~r 6 ~9 "si" 5 ~ = 3 3.3 2 9.3 5 81. 3 8 = 3 - .037037 - .000457 - .000009 - = 2.962497 + , Ans. RULE. Separate the given number into two parts, the first of whir// is the nearest perfect power of the same degree as the required root. Expand the result by the Binomial Theorem. Note. If the second term of the binomial is small, the terms in the expansion converge rapidly, and we obtain an approximate value of the required root by taking the sum of a few terms of the development. But if the second term is large, the terms converge slowly, and it requires the sum of many terms to insure a considerable degree of accuracy. EXAMPLES. Find the approximate values of the following to five deci- mal places : 3. #31. 5. #99. 7. #17. 4. #9. 6. #29. 8. #78. 328 ALGEBRA. XL. — SUMMATION OF INFINITE SERIES. 428. The Summation of a Series is the process of finding a finite expression equivalent to the series. Different series require different methods of summation, according to the nature of the series, or the law of its forma- tion. Methods of summing arithmetical and geometrical series have already been given (Arts. 369, 377, and 380). Methods applicable to other series will now be treated. RECURRING SERIES. 429. A Recurring Series is one in which each term, after some fixed term, bears a uniform relation to a fixed number of the preceding terms. Thus l + 2x + 3x 2 + Ax 3 + is a recurring series, in which each term, after the second, is equal to the product of the preceding term by 2 x, plus the product of the next term but one preceding by — x 2 . The sum of these constant multipliers is called the scale of relation of the series, and their coefficients constitute the scale of relation of the coefficients of the series. For example, in the series 1 + 2 x + 3 x' 2 + 4 x s + , the scale of relation is 2 x — x 2 , and the scale of relation of the coefficients is 2 — 1. 430. A recurring series is said to be of the first order when each term, commencing with the second, depends on the one immediately preceding; of the second order, when each term, commencing with the third, depends ujion the tiro im- mediately preceding ; and so on. If the series is of the first order, the scale of relation will consist of one term ; if of the second order, it will consist of two terms ; and, in general, the order and the number of terms in the scale of relation will correspond. 431. To find the scale of relation of the coefficients of a recurring series. SUMMATION OF INFINITE SERIES. 329 1. If the series is of the first order, it is a simple geometri- cal progression, and the scale of relation of the coefficients is found by dividing the coefficient of any term by the coefficient of the preceding term. 2. If the series is of the second order, let a, b, c, d, represent the consecutive coefficients of the series, and p + q their scale of relation. Then, c =p b + q a) d=p c + q b j (4> to determine p and q ; solving, we obtain ad — bc .. c 2 — b d p = 77 , and q = — — . ac — b 1 ac — b 1 3. If the series is of the third order, let a, b, c, d, e, f, represent the consecutive coefficients of the series, and p + q + r their scale of relation. Then, d =p c + q b + r a e =p d + q c + r b f=X> e + q d + r c from which we can find p, q, and ;■. 432. To ascertain the order of a series, we may first make trial of a scale of two terms, and if the result does not corre- spond with the series, we may try three terms, four terms, and so on, till the true scale of relation is found. If we assume the series to be of too high an order, the terms of the scale will take the form r. • 433. To find the sum of a recurring series, when the scale of relation of its coefficients is known. Let a + bx + cx 2 + dx 3 + +jx n ~ 3 + kx n ~ 2 + lx n ~ l + be a recurring se"ries of the second order. Let S denote the 330 ALGEBKA. sum of n terms of the series ; and let p + q be the scale of re- lation of the coefficients. Then, S= a + b x + c x 2 + d x 3 + + lx n ~ x p Sx=pax+pbx 2 +pcx 3 + + pkx n ~ 1 +plx n q Sx 2 = q ax 2 + q b X s + + qjx n ~ 1 +q k x n + qlx n + 1 Subtracting the last two equations from the first, S—p Sx — q Sx 2 = a + bx — pax— plx n — q kx n — qlx n + 1 the rest of the terms of the second member disappearing, be- cause, since p + q is the scale of relation of the coefficients, c =p b + q a, d =p c + q b, I =p k + qj. Therefore we have a + (b—p a) x — (p I + q k) x n — q I x n + 1 S = ■px — q x" the formula for finding the sum of n terms of a recurring series of the second order. But if n becomes indefinitely great, and the series is con- vergent, then the limiting values of the terms which involve x n and x n + ! must become 0, and we have at the limit s= a + (p-pa)x 1 — p x — qx 2 the formula for finding the sum of an infinite recurring series of the second order. If q = 0, then the series is of the first order, and conse* quently b =p a; then, 1 —px the formula for finding the stun of an infinite recurring series of the first order. (Compare Art. 380.) SUMMATION OF INFINITE SERIES. 331 In like manner, we should obtain a + (b — p a) x + (c —pb — qa)x 2 o — -z o 5 (o) 1 —p x — q x- — r x 6 w the formula for the summation of an infinite recurring series of the third order. 434. A recurring series, like other infinite series, originates from an irreducible fraction, called the generating fraction. The summation of the series, therefore, reproduces the frac- tion ; the operation being, in fact, the exact reverse of that in Art. 414. 435. 1. Find the sum of l + 2cc + 8ar+28z 3 +100x 4 + We must first determine the scale of relation of the coeffi- cients. In accordance with Art. 432, we first assume the series to be of the second order. We have a = i, b = 2, c==8, d = 28. Substituting in the values of p and q derived from (^4), Art. 431, we have p = 3 and q = 2. To ascertain if this is the proper scale of relation, consider the fifth term, 100 x 4 ; this should be 3 x times the preceding term, plus 2 x 2 times the next preceding term but one, or, 84 x 4 + 16 x*. This shows that the series is of the second order. Substituting in (1) the values of a, b, p, and q, we have l + (2 — 3)a; _ 1 — x b ~ i-Zx-2tf ~ l-3x-2x 2 ' EXAMPLES. Find the sum of the following series : 2. l + 2a- + 3x 2 + 5a; 3 + 8a; 4 + a ac a c 2 „ a c 3 3> b^¥ x + !F x "-l^ x+ 4. 4 + 9a; + 21ar+51;r 5 + 5. l + 3x + 5x 2 + 7x 3 + 332 ALGEBRA. 6. 2-a + 2a 2 -5a 3 +10a i -17a 5 + 7. 3 + 5x + 7 x' 2 + 13.x' 3 + 23 a; 4 + 45 x* + 8. l + 3x + 4:x 2 +7x 3 +llx 4 + 9. 2 + 4x-x*-3x s + 2x i + ±x 5 + DIFFERENTIAL METHOD. 436. The Differential Method is the process of finding any term, or the sum of any number of terms, of a regular series, by means of the successive differences of its terms. 437. If, in any series, we take the first term from the sec-, ond, the second from the third, the third from the fourth, and so on, the remainders will form a new series called the first order of differences. If the differences be taken in this new series in like manner, we obtain a series called the second order of differences ; and so on. Thus, if the given series is 1, 8, 27, 64, 125, 216, the successive orders of differences will be as follows : 1st order, 7, 19, 37, 61, 91, 2d order, 12, 18, 24, 30, 3d order, 6, 6, 6, 4th order, 0, 0, Hence, in this case there are only three orders of differences. 438. To find any term of a series. Let the series be a l) a 2> a 3> a i) a 5> a n1 a n+l> Then the first order of differences will be SUMMATION OF INFINITE SERIES. 333 the second order of differences will be a 3 — 2a 2 + a l , a A — 2a 3 + a 2 , a 5 — 2a i + a 3 , , the third order of differences will be a A — 3 a 3 + 3 a 2 — a x , a 5 — 3 a A + 3 a 3 — a 2 , , the fourth order of differences will be a 5 — 4 a A + 6 a& — 4 a 2 + a 1} , and so on ; where each difference, although a compound quan- tity, is called a term. Let now d x , d 2 , d s , d A , represent the first terms of the several orders of differences. Then, d l = a 2 — a l ; whence, a 2 = a l + d A . d 2 = a 3 — 2 a 2 + a x ; whence, a 3 — 2a 2 — a l +d 2 ^2ai + 2d 1 — a x + d 2 = a A + 2 d A + d 2 . d 3 = a A — 3 a s + 3 a 2 — a x ; whence, a A = a x + 3 d v + 3 d 2 -f- d s . d A = a 5 — A a A + 6 a 3 — Aa 2 +a l ; whence, a 5 = a l + 4^d 1 + 6d 2 + ±d 3 +d A . We observe that the coefficients of the value of a 2 are the same as the coefficients of the first power of a binomial; the coefficients of the value of a 3 are the same as the coefficients of the second power of a binomial ; and so on. Assume that this law holds for the nth term ; that is, that the coefficients of the value of a H are the same as the coefficients of the (n — l)th power of a binomial ; then, , „ . (n-l)(n-2) . a* = «i + (n - 1) (h + Q d 2 + („-!)(„- 2) <«-S) d3+ (1) If the law holds for the nth term in the given series, it will also hold for the «th term in the first order of differences ; or, 334 ALGEBRA. a n+1 -a n = d 1 +(n-l)d 2 + ^—-^f—^d ;i + (2) Adding (1) and (2), we have (n - 1) (n - 2) a , +1 = «!+[! + (n-l)]d l + (n - 1) + + (w-l)(w-2) (w-l)( w -2)(ra-3) ~|2- ~\T 2 ,/., = aj + n dx H r^- [2 + n — 2] d 2 (n - 1) (« - 2) ro , n(n-l) ? »(»-l)(n-2) . ,„ = «! + »!(/!+ -^ ~ <4 + ^ "^3 + (3) where the coefficients are the same as the coefficients of the rath power of a binomial. Hence, if the law holds for the nth term, it also holds for the {n + l)th term ; but we have shown it to hold for the fifth term, a 5 ; hence it holds for the sixth term ; and so on. That is, Formula (1) holds for any term in the series. When the differences finally become 0, the value of the nth term can be obtained exactly ; but, in other cases, the result is merely an approximate value. 439. To find the stem of any number of terms of a series. Let the series be a, b, c, d, e, (1) Let S denote the sum of the first n terms. Assume the series 0, a, a + b, a + b + c, a + b + c + d, (2) in which the (n + l)th term is obviously equal to the sum of n terms of the given scries ; that is, S is the (n + l)th term of series (2). Now the first order of differences of series (2) is SUMMATION OF INFINITE SERIES. 335 the same as series (1) ; hence, the second order of differences of series (2) is the same as the first order of (1) ; the third order of (2) is the same as the second order of (1) ; and so on. Then, letting a', d\, d' 2 , d' s , represent the first term, and the first terms of the several orders of differences of (2), we have a' = 0, d\ = a, d' 2 = d l ,d' s = d 2 , where a, d x , d 2 , are the first term, and the first terms of the several orders of differences of (1). But, by (3), Art. 438, the (n + l)th term of series (2) will be n (n - 1) „ n (n - 1) (n - 2) In this put for a', d\, d' 2 , d' 3 , their values; then n (n — 1) , n (n — 1) (re — 2) . S=na + — ^|2 — L d 1 +- ~" L d,+ (3) 440. 1. Find the 12th term of the series 2, 6, 12, 20, 30, The successive orders of differences will be as follows : 1st order, 4, 6, 8, 10, 2d order, 2, 2, 2, 3d order, 0, 0, Then a x = 2, d Y = 4, d 2 = 2, d s , d A , = 0, and n = 12. Substituting in (1), Art. 438, the 12th term (12 — 1) (12 — 2) = 2 + (12-l)4+ l >) } ; 2 = 2 + 44 + 110=156,^. 2. Find the sum of 8 terms of the series 2, 5, 10, 17, 1st order of differences, 3, 5, 7, 2d order of differences, 2, 2, 3d order of differences, 0, Then a = 2, d, = 3, d, = 2, n = S. 336 ALGEBRA. Substituting these values in (3), Art. 439, we have = 16 + 84 + 112 = 212, Ans. EXAMPLES. 3. Find the first term of the fifth order of differences of the series 6, 9, 17, 35, 63, 99, 4. Find the first term of the sixth order of differences of the series 3, 6, 11, 17, 24, 36, 50, 72, 5. Find the seventh term of the series 3, 5, 8, 12, 17, 6. Sum the first twelve terms of the series 1, 4, 10, 20, 35, 7. Sum the first hundred terms of the series 1, 2, 3, 4, 5, 8. Find the 15th term of the series l 2 , 2 2 , 3 2 , 4 2 , 9. Sum the first n terms of the series l 3 , 2 3 , 3 3 , 4 3 , 5 3 , 10. Sum the first n terms of the series 1, 2 4 , 3 4 , 4 4 , 5 4 , 6 4 , 11. If shot be piled in the shape of a pyramid, with a trian- gular base, each side of which exhibits 9 shot, find the number contained in the pile. 12. If shot be piled in the shape of a pyramid, with a square base, each side of which exhibits 25 shot, find the number contained in the pile. INTERPOLATION. 441. Interpolation is the process of introducing between terms of a series other terms conforming to the law of the series. SUMMATION OF INFINITE SERIES. 337 Its usual application is in finding intermediate numbers between tbose given in Mathematical Tables, which may be regarded as a series of equidistant terms. 442. The interpolation of any intermediate term in a series, is essentially finding the nth term of the series, by the differential method (Art. 438). Thus, Let t represent tbe term to be interpolated in a series of equidistant terms, and p the distance the term t is removed from the first term, a, expressed in intervals and fractions of an interval ; that is, p being the distance to the nth term, p = n — 1 intervals. In Formula (1), Art. 438, putting p for n — 1, the nth. term t=a+pdl+ pj£ ! =v. il+ p(*- 1 np-*> dt+ 443. 1. In the series :j^ , 7-7 > TE > T£ > T?? > , find the 13 14 15 16 17 middle term between 5-= and ^-7 . 15 lb Here, the first differences of the denominators are 1, 1, 1, 1, The second differences are 0, 0, 0, Whence, d t = 1, and d 2 = 0. 5 The distance to the required term is 2\ intervals, or p = ^- Make a = 13, the denominator of the first term ; then by the preceding formula, the denominator of the required term, 1 2 Therefore the required term is — or 757, Ans. 31 ol ~2 338 ALGEBRA. 2. Given ^94 = 9.69536, ^95 = 9.74679, y/ 96 = 9.79796 ; to find y/94i. Here, the first differences are .05143, .05117, and the second differences are -.00026, Whence, ^ = .05143, d 2 = -. 00026, . 1 . 1 The distance of the required term is - interval, or p = -? , Then the required term, 1(|-1 * = 9.69536 + 1 x .05143 + 4 ^ (- ,0( = 9.69536 + .01286 - J, (- .00026) + = 9.69536 + .01286 + .00002 + = (approximately) 9.70824, Ans. EXAMPLES. 3. Given ^64 = 4, ^65 = 4.0207, f 66 = 4.0412, ^67 = 4.0615 ; find ^66A 4. Given ^45 = 3.556893, ^47 = 3.608826, ^49 = 3.659306, ^51 = 3.708430; find ^48. 5. Given ^5 = 2.23607, yf 6 = 2.44949, ^7 = 2.64575, y/8 = 2.82843 ; find \'r>M. 6. (Jivcn the length of a degree of longitude in latitude 41°=4528 miles; in latitude 42° = 44.59 miles; in latitude 43°=43.8S miles; in latitude 44° = 43.16 miles. Find the length of a degree of longitude in latitude 11' 30'. LOGARITHMS. 339 7. If the amount of $ 1 at 7 per cent compound interest for 2 years is $ 1.145, for 3 years $ 1.225, for 4 years $ 1.311, and for 5 years $ 1.403, what is the amount for 4 years and 6 months ? XLI. — LOGARITHMS. 444. The logarithm of a quantity to any given base, is the exponent of the power to which the base must be raised to equal the quantity. For example, if a x = m, x is the exponent of the power to which the hase, a, must be raised to equal the quantity, m; or, x is the logarithm of m to the base a ; which is briefly expressed thus : x = log a m. 445. If a remain fixed, and m receive different values, a certain value of x Avill correspond to each value of m; and these values of x taken together constitute a System of Loga- rithms. And as the base, «., may have any value whatever, the number of possible systems is unlimited. For example, suppose a = 3. Then, since 3° = 1, by Art. 444, = log., 1 " 3 X = 3, " " l = log 3 3 " 3 2 =9, " " 2 = log 3 9 Hence, in the system whose base is 3, log 1 = 0, log 3 = 1, log 9 = 2, etc. Again, suppose a = 12. Then, since 12 1 = 12, 1 = log 12 12 " 12 2 = 144, 2 = log 12 144 Hence, in the system whose base is 12, log 12 = 1, log 144 = 2, etc. 340 ALGEBEA. 446. The only system in extensive use for numerical com- putations is the Common System or Briggs' System, whose base is 10. Therefore the definition of the common logarithm of a quantity is the exponent of that power of 1.0 which equals the quantity. Hence, Since 10° = 1, log 10 1 = 10 1 = 10, log 10 10 = 1 " 10 2 =100, log 10 100 = 2 TO 3 = 1000, log 10 1000 = 3 " 10- 1 = i = .l, log in .l = -l a 10- 2 = ^ = .01, log 10 .01=-2 '< 10- 3 -^ = .001, log 10 .001 = -3, etc. 447. It is customary in using common logarithms to omit the subscript 10 which denotes the base ; hence, we may write the results of Art. 446 as follows : log 1 = log .1=- 1 = 9- 10 log 10 = 1 log .01 = - 2 = 8 - 10 log 100 = 2 log .001 = -3 = 7 -10 log 1000 = 3 etc. The second form of the results in the second column will be found less complicated in the solution of examples. 448. We infer the following from the first column of Art. 447 : The logarithm of any number between 1 and 10, lies between and 1. The logarithm of any number between 10 and 100, lies be- tween 1 and 2. LOGARITHMS. 341 The logarithm of any number between 100 and 1000, lies be- tween 2 and 3, etc. Or, in other words, The logarithm of any number with one figure to the left of its decimal point, is equal to plus some decimal. The logarithm of any number with two figures to the left of its decimal point, is equal to 1 plus some decimal. The logarithm of any number with three figures to the left of its decimal point, is equal to 2 plus some decimal, etc. 449. Reasoning in the same way from the second column of Art. 447, The logarithm of any number between 1 and .1, lies between and 9 — 10, or between 10 — 10 and 9 — 10. The logarithm of any number between .1 and .01, lies be- tween 9 — 10 and 8 — 10. The logarithm of any number between .01 and .001, lies be- tween 8 — 10 and 7 — 10, etc. Or, in other words, The logarithm of any decimal with no zeros between its point and first figure, is equal to 9 plus some decimal — 10. The logarithm of any decimal with one zero between its point and first figure, is equal to 8 plus some decimal — 10. The logarithm of any decimal with two zeros between its point and first figure, is equal to 7 plus some decimal — 10, etc. 450. It will be seen from the two preceding articles that in general the logarithm of a number consists of two parts, one integral, the other decimal. The integral part is called the characteristic ; the decimal part, the mantissa. For rea- sons which will be given hereafter, only the mantissa of the logarithm is given in the tables ; the characteristic must be supplied by the reader. The rules for characteristic are based on the results obtained in the last parts of Arts. 448 and 449. 342 ALGEBRA. 451. I. If the number is greater than 1, the characteristic is 1 less than the number of figures to the left of the decimal [lit'; lit. For example, characteristic of log 354.89 = 2, characteristic of log 906328.3 = 5, etc. II. If fin' number is less than 1, the characteristic is found by subtracting the number of zeros between the decimal point cut/ first significant figure from 9; writing — 10 after the mantissa. For example, characteristic of log .00792 = 7, with — 10 after the mantissa; characteristic of log .2583 = 9, with —10 after the mantissa ; etc. It is customary in ordinary computation to omit the — 10 after the mantissa ; it should he remembered, however, that it is really a part of the logarithm, and should be allowed for, and subjected to precisely the same operations as the rest of the logarithm. Beginners will find it useful to write it in all cases ; and in some problems it cannot conveniently be omitted. Note. Many writers, in dealing with the characteristics of the loga- rithms of numbers less than 1, combine the two portions of the characteris- tic, writing the result as a negative characteristic before the mantissa. Thus, instead of such an expression as 7.603582-10, the student will fre- quently find 3.6035S2 ; a minus sign being written over the characteristic, to denote that it alone is negative, the mantissa being always positive. The objection to this notation is the inconvenience of using numbers partly positive and partly negative. PROPERTIES OF LOGARITHMS. 452. In any system the logarithm of unity is zero. For, since a = 1, for any value of a, = log a 1. 453. In any system the logarithm of the base itself is unity. For, since a 1 = a, for any value of a, 1 = log„ a. LOGARITHMS. 343 454. hi any system, whose base is greater than unity, the logarithm of zero is minus infinity. For, since or 00 = — - = — = 0, — cc = log a 0. a go If the base is less than unity, the logarithm of is + cc . 455. In any system the logarithm of the product of any number of factors is equal to the sum of the logarithms of those factors. Assume the equations, - "H whence, by Art. 444, [ x ~~ !° g " Multiplying, a x X cC> — m n, or a x + \ — m n Whence, x + y = log a m n Substituting values of x and y, log a m n — log a m + log a n. If there are three factors, m, n, andj?, log a m np = log a (m n Xp) = (Art. 455) log a m n + \og a p = log a m + log a n + log a p. An extension of this 'method will prove the theorem for any number of factors. By the application of this theorem, we may find the loga- rithm of a number, provided we know the logarithm of each of its factors. For example, given log 2 = 0.301030, log 3 = 0.477121, required log 72. log 72 = log (2 x 2 x 2 x 3 X 3) = log 2 + log 2 + log 2 + log 3 + log 3 = 3 x log 2 + 2 x log 3 = 0.903090 + 0.954242 = 1.857332, Ans. 344 ALGEBRA. EXAMPLES. Given log 2 = 0.301030, log 3 = 0.477121, log 7 = 0.845098, calculate : • 1. log 48. 4. log 98. 7. log 1G8. 10. log 3087. 2. log 441. 5. log 84. 8. log 7056. 11. log 15552. 3. log 56. 6. log 567. 9. log 504. 12. log 14406. 456. In any system the logarithm of a fraction is equal to the logarithm of the numerator minus the logarithm of the denominator. Assume the equations, a x = m) i f x = log a m „ > whence, { , oa ay — n J (y = \og a n Dividing, ar m , m — = — , or a x ~ v = — a^ n n Whence, x-y = \og a -- Substituting values of x and y, loga — = log a m — log a n, IV By this theorem, a logarithm being given, we may derive certain others from it. For instance, if we -know log 2 = 0.301030, then log 5 = log ^ = i g io - log 2 = 1. - 0.301030 = 0.698970. 41 EXAMPLES. Given log 2 = 0.301030, log 3 = 0.477121, log 7 = 0.845098, calculate : LOGARITHMS. 345 1. log 15. 4. log 175. 7. logTf 2. log 125. 5. log 3i. 8. log—, 10 T 3. log—. 6. loglH. 9. log5£. 457. In any system the logarithm of any power of a quantity is equal to the logarithm of the quantity, multiplied by the exponent of the power. Assume the equation, a x = in, whence, x = log a m Eaising both members of the assumed equation to the^th power, (a x )P = mP, or aP x = mP Whence, px = log„ mP Substituting the value of x, log a mP=p\og a m. 458. In any system the logarithm of any root of a quan- tity is equal to the logarithm of the quantity, divided by the index of the root. For, log a v' m = log a (m7) = (Art. 457) - log a m. 459. In the common system, the mantissa; of the loga- rithms of all numbers having the same sequence of figures will be the same. For example, suppose we know that log 3.053 = .484727. Then,log30.53=log(3.053xl0)=log3.053+logl0=.484727 + 1 = 1.484727. Also, log 30530 = log (3.053 x 10000) = log 3.053 + log 10000 = .484727 + 4 = 4.484727. 346 ALGEBRA. Again, log.03053=log (^~J=log3.053-logl00=.484727 -2= .484727 + 8-10 = 8.484727-10. It is clear, then, that if a number he multiplied or divided hy any integral power of 10, thereby producing another number having the same sequence of figures, the mantissse of their logarithms will be the same. Or, to illustrate, if log 3.053 = .484727, then, log 30.53 = 1.484727 log .3053 = 9.48472? - 10 log 305.3 = 2.484727 log .03053 = 8.484727 - 10 log 3053. = 3.484727 log .003053 = 7.484727 - 10 etc. etc. We may now see the reason why, as stated in Art. 450, only the mantissa? are given in the table ; for if we wish to find the logarithm of any number, we have only to find the mantissa of the sequence of figures composing it from the table, and can prefix the proper characteristic, depending on the position of the decimal point, in accordance with the rules stated in Art. 451. This property of logarithms is only enjoyed by the com- mon system, and constitutes its superiority over all others. 460. Given the logarithm, of a quantity to a certain base, to calculate the logarithm of the same quantity to any other base. Assume the equations, a x = m) i (x = log, m ,,. > whence, < , oa O'J =m) ' \y = log 6 m From the assumed equations, a x = by l I £ Hence, (a*)?/ = (tity, or av = b Whence, - = lo£„ b y X °h y l°g a h LOGARITHMS. 347 Substituting the values of x and y, , log„ m log,, m = . . log„£ That is, if we know the logarithm of m to a certain base, a, its logarithm to any other base, b, is found by dividing by the logarithm of b to the base a. 461. To show that log a 6 X log 6 a = l, for any values oj a and b. Assume the equation, a x = b, whence x = log a b Taking the - power of both members, 00 III (a x y = b x , or b x = a Whence, - = log 6 a 7 x Therefore, log a b X log 6 a = x X - = 1- 462. We append a few examples to illustrate the applica- tions of Arts. 455, 456, 457, and 458. e L l0g © d= 4 l0g !' (Art. 457) = - (log a - log b), (Art. 45G) . Co 2. log % a *V b = log Q a x m \J b) - log % e, (Art. 456) = log \f a + log y 7 ^ — log y/ c, (Art. 455) = - log a-\ log b log c, (Art. 458). n m P The following are proposed as exercises a b c ~d~e. 3. log -7— = log a + log b + log c — log d — log e. 348 ALGKBBA. 4. log (J/«xJ 3 X c-) = - log « + 3 log & + ^ log a. 5. log^ = r, log2--log3. 3 g o o 6. log \7 — = - (2 log a — log b — log c). V c n „ . V a b 1 „ _ _ . 1 7 - log -^y— = - (log « + log &) - — log c. 8 - l°g > = t log a — log & — s log c — 2 log tf. bc$d 2 4 ° ( s I a _ ™ \ 1 to 9 - lo S VV ^ + ( c d ) "J = 5 ( lo S « - log ^) + — (logc+ logd)- USE OF THE TABLE. 463. The table (Appendix) gives the mantissa? of the logarithms of all numbers from 1 to 10000, calculated to six decimal places. On the first page of the table are the loga- rithms of the numbers between 1 and 100. This table is added simply for convenience, as the same mantissae are to be found in the rest of the table. To find the logarithm of any member consisting of four figures. Find, in the column headed N, the first three figures of the given number. Then the mantissa, (if the required logarithm will be found in the horizontal line corresponding, in the ver- tical column which lias the fourth figure of the given number at the top. If only the last four figures of the mantissa are found, the first two figures may be obtained from the nearest mantissa above, in the same vertical column, which consists of six figures. Finally, prefix the proper characteristic (Art. 451). LOGARITHMS. 349 For example, log 140.8 = 2.148603 log .05837 = 8.766190 - 10 log 8516. = 3.930236 For a number consisting of one or two figures, use the first page of the table, which needs no explanation ; for a number of three figures, look in the column headed 1ST, and take the mantissa corresponding in tbe column headed 0. For exam- ple, log 94.6 = 1.975891. 464. To find the logarithm of a number of more than four figures. For example, let it be required to find log 3296.78. From the table, we find log 3296 = 3.517987 log 3297 = 3.518119 That is, an increase of one unit in the number produces an increase of .000132 in the logarithm. Then' evidently an in- crease of .78 unit in the number will produce an increase of .78 X .000132 in the logarithm = .000103 to the nearest sixth decimal place; Therefore, log 3296.78 = log 3296 + .000103 = 3.517987 + .000103 = 3.518090, Ans. Note. The foregoing method is based upon the assumption that the differences of logarithms are proportional to the differences of their corre- sponding numbers, which is not strictly correct, but is sufficiently exact for practical purposes. We derive the following rule from the above operation : Find in the table the mantissa of the first four figures, without regard to tin' position of the decimal point. Find the difference between this and the mantissa of the next 7iigher number of four figures ; (called the tabular dif- ference, and to be found in the column headed D on each page : see Note on page 350.) Multiply the tubular difference by the rest of the figures of the given number, with a decimal point before them. Add the result to the mantissa of the first four figures. Prefix the proper characteristic. 350 ALGEBRA. 1. Find the logarithm of .02243076. Mantissa of 2243 = 350829 Tabular difference = 194 lj> .076 350844 1.164 13.58 Correction = 14.744 = 15 nearly. Am. 8.350844-10. Note. To find the tabular difference mentally, subtract the last figure of the mantissa from the last figure of the next larger, and take the nean si whole number ending in that figure to the number in the column headed D in the same line. For instance, in finding log .02243076, the last figure of the mantissa of 2243 is 9, and of the next larger mantissa, 3 ; 9 from 13 leaves 4, and the nearest number ending in 4 to 193, the number in the column headed D, is 194, the proper tabular difference. EXAMPLES. Find the logarithms of the following numbers : 10 < "8 oa "" 1 .110 \ji. "'^ ^""tTlMg XX LLJ_U.k/^>Xi z> ■ 2. .053. 6. 33.0908. 10. 912.2:..".. 3. 51.8. 7. .0002851. 11. .870092. 4. .2956. 8. 65000.63. 12. 7303. Oi 8 5. 1.0274. 9. .001030741. 13. .0436927 14. Given log 7.83 = .89376, log 7.84 = .89432; find log 78309. 15. Given log .05229 = 8.718419-10, log .05230 = 8.718502 -10; find log 52.2938. 16 Given log 315.08 = 2.4984208, log 315.09 =2.4984346; find log .003150823. 17. Given log 18.84 = 1.275081, log 18.87 = 1.275772 ; find log .188527. 18. Given log 9.5338 = .9792660, log 9.5342 = .9792843 ; find log 9534071. LOGARITHMS. 351 465. To find the number corresponding to a logarithm. For example, let it be required to find the number whose logarithm is 3.693845. Since the characteristic depends only on the position of the decimal point, and in no way affects the sequence of figures corresponding, we ought to obtain all of the number corre- sponding, except the decimal point, by considering the man- tissa only. We find in the table the mantissa 693815, of which the corresponding number is 4941, and the mantissa 693903, of which the corresponding number is 4 ( .)42. Tbat is, an increase of 88 in the mantissa produces an in- crease of one unit in the number corresponding. Hence, an increase of 30 in the mantissa will produce an increase of §§ of a unit in the number, or .34 nearly. Therefore, Number corresponding = 4941 + -34 = 4941.34, Ans. We base the following rule on the above operation : Find in the table the next less mantissa, the four figures corresponding, and the tabular difference. Subtract the next less mantissa from the given, mantissa. Divide the remainder bg the tabular difference / (the quo- tient in general cannot be depended upon to more than two decimal places.) Annex all of the quotient except the decimal point to the first four figures of the number. Point off. Note. The rules for pointing off are the reverse of tire rules for charac- teristic given in Art. 451 : I. If — 10 is not written after the mantissa, add 1 to the characteristic, giving the number of figures to the left of the decimal point. II. If — 10 is written after the mantissa, subtract the characteynstic from 9 ; giving the number of zeros to be placed between the decimal point and first figure. 352 ALGEBRA. 1. Find the number whose logarithm is 7.950185 — 10. 950185 Next less mantissa =950170; four figures corresponding =8916. Tabular difference =49) 15.00 (.31 nearly. 147 ~30 Therefore, number corresponding = .00891631, Ans. EXAMPLES. Find the numbers corresponding to the following : 2. 1.880814. 6. 8.044891-10. 10. 0.990191. 3. 9.470410-10. 7. 2.270293. 11. 7.115658-10. 4. 0.820204. 8. 9350064-10. 12. 8.535003-10. 5. 4.745126. 9. 3.000027. 13. 1.670180. 14. Given log 113 = 2.05308, log 114 = 2.05690 ; find num- ber corresponding to 1.05411. 15. Given log .08630 = 8.936011 - 10, log .08631 = 8.936061 — 10 ; find number corresponding to 0.936049. 16. Given log 2.0702 = .3160123, log 2.0703 = .3160333 ; find number corresponding to 9.3160138 — 10. 17. Given log 548 3 = 2.739018, log 548.9 = 2.739493 ; find number corresponding to 7.739416 — 10. 18. Given log 7.3488 = .8662164, log 7.3492 = .8662401 ; find number corresponding to 2.8662350. 466. In the application of Arts. 455, 456, 457, and 458, we have to perform the operations of Addition, Subtraction, Mul- tiplication, and Division with logarithms. As some of the problems which may arise arc peculiar, wo give a few hints as to their solution, which will be found of service. 1. Addition. If, in the sum, — 10, —20, —30, etc., are written after the mantissa, and the characteristic standing be- LOGARITHMS. 353 fore the mantissa is greater than 9, subtract from both parts of the logarithm such a multiple of 10 as will make the charac- teristic before the mantissa less than 10. For example, 13.354802 - 10 should be changed to 3.354802 ; 28.964316 - 30 should be changed to 8.964316 - 10 ; etc. 2. Subtraction. In subtracting a larger logarithm from a smaller, or in subtracting a negative logarithm from a posi- tive, the characteristic of the minuend should be increased by 10, — 10 being written after the mantissa to compensate. For example, to subtract 3.121468 from 2.503964, we write the minuend in the form 12.503964 — 10 ; subtracting from this 3.121468, we have as a result 9.382496 — 10. To subtract 9.635321 — 10 from 9.583427 - 10, we write the minuend in the form 19.583427 — 20 ; subtracting from this 9.635321 - 10, we have as a result 9.948106 - 10. 3. Multiplication. The hint already given for reducing the result of Addition, applies with equal force to Multiplication. To multiply a logarithm by a fraction, multiply first b} r the numerator, and divide the result by the denominator. 4. Division. In dividing a negative logarithm, add to both parts of the logarithm such a multiple of 10 as will make the quantity after the mantissa exactly divisible by the divisor, with — 10 as the quotient. For example, to divide 7.402938 — 10 by 6, we add 50 to both parts of the logarithm, giving 57.402938 — 60. Dividing this by 6, we have as a result 9.567156 — 10. EXAMPLES. 1. Add 9.096004 - 10, 4.581726, and 8.447510 - 10. 2. Add 7.196070 - 10, 8.822209 - 10, and 2.205683. 3. Subtract 0.659321 from 0.511490. 4. Subtract 7.901338 - 10 from 1.009800; 5. Subtract 9.156243 - 10 from 8.750404 - 10. 354 ALGEBRA. 6. Multiply 9.105107 - 10 by 3. 7. Divide 8.452G33 - 10 by 4. 8. Divide 9.670392 - 10 by 11. 9. Multiply 9.6G8311 - 10 by ?. SOLUTIONS OF ARITHMETICAL PROBLEMS BY LOGARITHMS. 467. In finding the value of any arithmetical quantity by logarithms, we first find the logarithm of the quantity, as in Art. 462, by the aid of the table, and then find the number corresponding to the result. 1. Find the value of .0631 X 7.208 X 512.72. By Art. 455, log (.0631 x 7.208 x 512.72) = log .0631 + log 7.208 + log 512.72 log .0631= 8.800029-10 log 7.208= 0.857815 log 512.72= 2.709880 Adding, .-. log of Ans. = 12.367724 - 10 = 2.367724 (Art. 466, 1) Number corresponding to 2.367724 = 233.197, Ans. „. , „ . . 3368.52 2. Find the value of -^^g. log HJIJH = log 3368.52 - log 7980.04 log 3368.52 = 13.527439 - 10 (Art. 466, 2) log 7980.04= 3.902005 Subtracting, .-. log of Ans. = 9.625434—10 Number corresponding =.422118, Ans. LOGARITHMS. 355 3. Find the value of (.0980937) 5 . log (.0980937) 5 = 5 x log .0980937 log .0980937 = 8.991641 - 10 5 Multiplying, .-. log of Ans. = 44.958205 - 50 = 4.958205-10 Number corresponding = .0000090825, Ans. 4. Find the value of ^2.36015. log ^ 2.3601 5 = * log 2.36015 log 2.36015 = 0.372940 Dividing by 7, .-. log of Ans. = 0.053277 Number corresponding = 1.13052, Ans. 2 v^5 5. Find the value of — — . 3* log $ - ] °g 2 + I log 5-5 log 3 o log 2 = 0.301030 log 5 = 0.698970 ; divide by 3 = 0.232990 log 3 = 0.477121 0.534020 Multiply by 5, = 2.385605 ; divide by 6 = 0.397601 Subtracting, .-. log of Ans. = 0.136419 • Number corresponding = 1.36905, Ans. Note. The work of the next two examples will be exhibited in the customary form, the — 10's being omitted after the mantissse. See Art. 451. 6. Find the value of ^.00003591. 356 ALGEBRA. log {/ .00003591 = ^ log .00003591 log .00003591 = 5.555215 7)5.555215 log of Ans. Z 9.365031 (Art. 466, 4) Ans. = .231756. m -n- ", i i r // -032956 \ 7. Find the value of W ( - 7.96183/' los \J (^SiH) = \ ^ - 03295G " log 7,96183) log .032956 = 8.517934 log 7.96183 = 0.901013 2)7.616921 log of Ans. = 8.808460 Ans. = .0643369. Note. In computations by logarithms, negative quantities are used as if they were positive ; the sign of the result being determined irrespective of the logarithmic work. EXAMPLES. 468. Calculate; by logarithms, the values of the following : 1. 9.23841 x .00369822. 5. ^3. * 3.70963 x 286.512 g ,g 1633.72 * ' V 3. (23.846-t) 8 . 7. ^5. 4. (- .0009296S7)*. 8. ^.0042937. LOGARITHMS. 357 18 9. V- 6829.586. 112 10. (1.05624) 11. (- .0020001G)i£. 12. 2? x (- 3)*. 13. 14. 3 5 T (-2)* 3^ (- 4) § 15. m ii 16. V 7239.812. 17. V .00230508. 19. 35 113 / .0872635 U \ .132088 / " «. i/\- 22. i> 23 21 13* 24. f2x('3xf4. // 3258.826 \ V V 49309.8 ) ' 25 / - 31.6259 W ' V 429.0162 27 _ (625.343)- (.732465) t 28. 29. 30. V .000128883 y. 000827606* (_ .746892) ^ - (.234521)^ ty .00730007 " * (.682913) ^ 18. V- .000009506694. 31. y 5.95463 x V 61.1998 V 298.5434 32. (538.217 x .000596899)^. 33. - 304.698 x .9026137 .00776129 X- 16923.24 34. (18.9503) 11 x (-.213675) 14 . 358 ALGEBRA. A Orro I Q 35. V 3734.89 x .00001108184. 36. (2.03172)* x (.712719)* y- .00819323 x (.0628513) * - .9834171 " 6/^T7T7777^T- .8/ 38. \/.035 x V .626671 X V-M721033. EXPONENTIAL EQUATIONS. 469. An Exponential Equation is one in which the un- known quantity occurs as an exponent. To solve an equation of this form, take the logarithms of hoth members according to Art. 457'; the result will be an equation which can be solved by ordinary algebraic methods. 1. Given 31* = 23 ; find the value of x. Taking the logarithms of both members, log (31*) = log 23 or, by Art. 457, x log 31 = log 23 mi lo S 23 1-361728 MWVr - , Whence, X = ^ = -^—^ = .91307 < , Ans. The value of the fraction '.^~^ may be obtained bv di- 1.491362 J J vision, or better by logarithms, as in Art. 468. 2. Given .2* = 3 ; find the value of x. Taking the logarithms of both members, x log .2 = log 3 log 3 .477121 .477121 \Y hence, x = ' log .2 9.301030 — 10 .698970 We may find the value of the fraction by logarithms exactly as if it were positive, and prefix a — sign to the result. Thus. LOGARITHMS. 359 log .477121 = 9.678628 - 10. log .698970 = 9.814458 - 10 Subtracting, = 9.834170 - 10 Number corresponding = .682606 Therefore, x = — .682606, Ans. EXAMPLES. Solve the following equations : 3. II 1 = 3. 5. 13* = .281. 7. 5*~ 8 = 8 2 * +1 . 4. .3 r = .S. 6. .703* = 1.09604. 8. 23 3 * + 5 = 31 2 *- 3 . APPLICATION OF LOGARITHMS TO PROBLEMS IN COMPOUND INTEREST. 470. Let P = the principal, expressed in dollars. Let t = the interval of time during which simple interest is calculated, expressed in years and fractions of a }*ear. For instance, if the interest is compounded annually, t = 1 ; if semi-annually, t = - ; etc. Let P = the interest of one dollar for the time t. Let n = the number of years. Let Ai, A 2 , A 3 , be the amounts at the ends of the 1st, 2d, 3d, intervals. Let A be the amount at the end of n years. Then A l = P + PP = P(l + P) A 2 = A 1 + A 1 B = A 1 (1 + R) =p (i + p) (i + p) = p (i + py A 3 = A, + A 2 P = A 2 (1 + P) = p(i + py 2 (i + p) = p(i + py 300 ALGEBRA. 71 As there are - intervals, the amount at the end of the last, v according to the law observed above, A =* P (1 + Sp. 1. Given P, t, R, and n, to find A. n As A = P (1 + R) *., we have by logarithms, log A = log P (1 + Rp = log P + log (1 + Rp 71 = log P + - log (1 + R). h Example. What will be the amount of $7,325.67 for 3 years 9 months at 7 per cent compound interest, the interest being compounded quarterly ? Here P = 7325.67, t = j , R — .0175, n = 3f, - == 15. log P = 3.864848 i log (1 + R) = 0.007534 ; multiply by 15 = 0.113010 Adding, . •. log of A = 3.977858 Number corresponding, A = $ 9502.93, Ans. 2. Given t, R, n, and A, to find P. n A As A = P (1 + R)t , .-. P = ; or, by logarithms, (1 + R)~ /log P = log A — log (1 + Rp = log A — " log (1 + R). Example. What sum of money will amount to $ 1 ,76« \.5B at 5 per cent compound interest in 3 years, the interest being compounded semi-annually ? LOGARITHMS. 361 1 n Here t = % , B = .025, » = 3, A = 1763.55, - = 6. log ^ = 3.246388 log (1 + P) = 0.010724; multiply by 6 = 0.064344 Subtracting, .-. log P = 3.182044 Number corresponding = f> 1520.70, Ans. 3. Given P, t, P, and A, to find n. In Art. 470, 1, we sbowed that log^ = logP + -log(l + P) 1/ .•.^log(l + P)=log^-logP £ (log A — log P) * • l ~ log (1 + 5) • Example. In bow many years will $300.00 amount to 8 400.00 at 6 per cent compound interest, the interest being compounded quarterly ? Here P = 300, t = | , P = .015, A = 400. log 400 - log 300 2.602060 - 2.477121 .124039 .'.71 = 4 log 1.015 ~ 4 x .006466 .025864 = 4.83 years, Ans. 4. Given P, t, n, and A, to find P. n We sbowed, in Art. 470, 3, that - log(l + P) = log^-logP V 1 /1 , T>\ log^ — logP ••• log (1 + P) ■ • Example. If $ 500.00 at compound interest amounts to $689.26 in 6 years and 6 months, the interest being com- pounded semi-annually, what is tbe rate per cent per annum '? G2 ALGEBRA. 1 n Here P = 500, t = 7> ,n = 6h,A = 689.26, - = 13. z t , ,- ™ log 689.26- log 500 13 log 689.26 = 2.838383 log 500 = 2.698970 Subtracting, = 0.139413 Dividing by 13, .-. log (1 + B) = 0.010724 Number corresponding = 1.025 = 1 + II, or R = .025. That is, one dollar gains $ .025 semi-annually ; or the rate is 5 per cent per annum. EXPONENTIAL AND LOGARITHMIC SERIES. 471. We know that for any values of n and x, 1 + x / 1 \ n = ( 1 + n, Expanding by the Binomial Theorem, we obtain 1 n (n - 1) 1 n (n - 1) (n - 2) 1 1 + n - H r?i 5 H rr; ? + w w." [3 ?r . 1 v x (nr—1) 1 ?? x(nx— V)(nx— 2) 1 w 2 ?r 3 ?r + or, !_! (l-l)(!- 2 ) T77- T^ + L3 x \x x \x V n I n = 1 + x H r^ + 11 \1 ; + LOGARITHMS. 363 This is true for all values of n ; hence, it is true however large n may he. Suppose n to he indefinitely increased. Then 1 2 the limiting values of the fractions - , — , etc., are (Art. 210). n n Hence, at the limit, we have, 1 1 1 + 1 + 777 + T7V + -f^ + Sr + [3_ J ' ' [2 ' [3 The series in the bracket we denote by e ; hence, 2 3 x x <? = ! + * + - + - + 472. To expand a x in poivers of x. Let a = e m ; whence (Art. 444), 7ft = log c ft. m 2 a? 2 ra 3 x 3 Then a* = e ra x = (Art. 471) 1 + m a; + -y^- + -r^- + Substituting the value of m, V 2 /I \ *\ a* = 1 + (l0g e ft) X + (log, ft) 2 — + (log e ft) 3 j^ + This result is called the Exponential Theorem. 473. The system of logarithms which has e for its base, is called the Napierian System, from Napier, the inventor of logarithms. The value of e may be easily calculated from the series of Art. 471, and will be found to be 2.7182818 474. To expand log,, (1 + x) in poivers ofx. a* = {l+(a-l)}* = l + x(a l -l).+ X(x ~ 1) (a-iy + g (s-l)(s-2) + = l + x {(ft- 1) - - ( -^ + ^^ } + terms con- taining x 2 , x s } etc. 364 ALGEBRA. But (Art. 472), a x = 1 + x (log e a) + terms containing x% As the two values of a x are equal for all values of x, by the Theorem of Undetermined Coefficients the coefficients of x in the two expressions are equal; hence, log e a=(a — l) ^ 1 3 Putting a = 1 + x, and therefore a — 1 = x, we obtain Syt" /ytO log e (l + «)=» — — + — — Note. This formula might be used to calculate Napierian logarithms ; but unless x is a very small fraction, the series in the second number is either divergent or converges very slowly, and hence is useless in most cases. 475. To obtain a more convenient formula for calculating the Napierian logarithm of a number. X 2 X s X* X 5 By Art. 4,4, log e (1 + x) = x — y + y — y + y — put X = — X, X 2 X 3 X* X s .-.log e (l — x)=—x Subtracting, 2 x 3 2x h • . log e (1 + x) - log, (l-x) = 2x+ — +- ir + or, by Art. 456, log,, f j 1 + x\ c ( x 3 x 5 > = 2 [x+- 7r +- T r + Let x = x) 3 5 1 1 + 2n + l 1 l+x_ 2n + l _ 2n + 1 + 1 _ 2n + 2 __ n + l l—x~~7 ~T~ ~2n + l — l~ 2 u n 2n + l LOGARITHMS. 365 Substituting, .\ log, p— J = lo ge (n + 1) — log e » ~~ \2n + 1^3(2n + iy ^ 5 (2 n + l) 3 T " / ■■■^g e (n+l)=\o ge n+2(^ + 3 ^ +1) ,+ 5{:Jn+1)b + ) 476. To calculate log e 2, put n = 1 in the formula of Art. 475. / 1 1 1 \ ...log e 2 = log e H-2^2 TI +3 (2 + 1) s+5^2q:-i)5+ j or, since log e 1 = 0, /l 1 1 1 1 1 log. 2 = 2 ^3 + 81 + 12i5 + 15309 + 177147 + 1948617" 1 """ = 2 (.3333333 + .0123457 + .0008230 + .0000653 + .0000056 + .0000005 + ) = 2 x .3405734 = .6931468 = .693147, correct to the sixth decimal place. From log,, 2, we may calculate log e 3 ; and so on. We shall find log e 10 = 2.302585. 477. Tn calculate the common logarithm of a number from its Napierian logarithm. By Art. 460, changing b to 10, and a to e, we obtain "»■ " = feS = &3S3«i b ~ » = - 434 - 0945 X ,0g ' * For instance, log I0 2 = .4342945 x .693147 = .301030. The multiplier by which logarithms of any system are de- rived from the Napierian system, is called the modulus of that system. Hence, .4342945 is the modulus of the common sys- tem. 366 ALGEBRA. As tables of common logarithms are met with more fre- quently than tables of Napierian, a rule for changing common logarithms into Napierian may be found convenient. RULE. Divide the common logarithm by .4342945. For example, to find the Napierian logarithm of 586.324, common log 586.324 = 2.768138 Divide by .4342945, .-. Napierian log 586.324 = 6.373873, Ans. Another method would be to multiply the common logarithm by 2.302585, the reciprocal of .4342945. Napierian logarithms are sometimes called hyperbolic loga- rithms, from having been originally derived from the hyper- bola. They are also sometimes called natural logarithms, from being those which occur first in the investigation of a method of calculating logarithms. Napierian logarithms are seldom used in computation, but occur frequently in theoretical investigations. ARITHMETICAL COMPLEMENT. 478. The Arithmetical Complement of the logarithm of any quantity is the logarithm of the reciprocal of that quantity. For example, if log 4098 = 3.612572, then ar. co. l#g 4098 = log — — = log 1 - log 4098 ° 4098 ° ° = - 3.612572 = 6.387428 - 10. Again, if log .06689 = 8.825361 - 10, then ar. co. log .06689 = log --?— - = - (8.825361 - 10) = 10 - 8.825361 = 1.174639. LOGARITHMS. 367 The following rules will be evident from the preceding illustrations : To find the arithmetical complement of a positive loga- rithm, suit nut it from 10, writing — 10 after the mantissa. To find the arithmetical complement of a negative loga- rithm, subtract that portion of it besides the — 10 from 10. The only application of this is to exhibit the work of calcu- lation by logarithms in a more compact form in certain cases. It depends on the' principle that subtracting a logarithm or adding its arithmetical complement gives the same result. For, suppose we are to calculate by logarithms. . a X b . ( . 1 1 l0 «7^ = l0g l ax( ' x c x ( z = log a + log b + log - + log 1 i 1 c + ] ° S d = log a -f- log b + ar. co. log c + ar. co. log d. That is, the work can be exhibited in the form of the addi- tion of four logarithms, instead of the subtraction of the sum of two logarithms from the sum of two others. The principle is only applicable to the case of fractions ; and the rule to be used is, Add together the logarithms of the quantities in the numer- ator, and the arithmetical complements of the logarithms of the- quantities in the denominator. Example. Calculate the value of „ 't, — _1 00 . 1 613.8 x .0* .23 log Qi^xljfn = l0 S 79 ' 23 + lQ g 10 - 39 + ar " co - Io S 613 - 8 + ar. co. log .07723 368 ALGEBRA. log 79.23= 1.898890 log 10.39 = 1.016616 ar. co. log G13.8 = 7.211973 - 10 ar. co. log .07723= 1.112211 Adding, .-.log of Ans. = 11.239693 - 10 = 1.239693 Number corresponding = 17.3657, Ans. Note. The arithmetical complement may be calculated mentally from the logarithm, by subtracting the last significant figure from 10, and all the others from 9. MISCELLANEOUS EXAMPLES. 479. 1. Find log 3 2187. (See Art. 111.) 2. Find log 5 15625. 3. Find the logarithm of -rr to the base —2. * 61 4. Find the logarithm of — to the base 8. 5. Find the characteristic of log 2 183. 6. Find the characteristic of log 5 4203. 7. Given log 2 = .301030, how many digits are there in 2 17 ? 8. Given log 3 = .477121, how many digits are there in 3^? 9. Findlog 13 56. (See Art. 460.) 10. Find log 8 163. 11. Find loggo 411. 12. What sum of money will amount to § 8705.50, in 7 years, at 7 per cent compound interest, the interest being com- pounded annually ? GENERAL THEORY OF EQUATIONS. 360 13. In how many yours will a sum of money double itself at 6 per cent compound interest, the interest being compounded semi-annually ? 14. What will be the amount of $1000.00 for 38 years 3 months, at 6 per cent compound interest, the interest being compounded quarterly ? 15. At what rate per cent per annum will $2500.00 amount to $ 3187.29 in 3 years and 6 months, the interest being com- pounded quarterly ? 16. In bow many years will 8 9681.32 amount to $ 15308.70 at 5 per cent compound interest, the interest being compounded annually ? 17. Using the table of common logarithms, find the Na- pierian logarithm of 52.9381 (Art. 477). 18. Find the Napierian logarithm of 1325.07. 19 Find the Napierian logarithm of .085623. 20. Find the Napierian logarithm of .342977. XLII. — GENERAL THEORY OF EQUATIONS. 480. The general form of a complete equation of the nth degree is x n -\- 1~> x n ~ l + q x n ~' 2 + + t x 2 + u x + v = "Where n is a positive integer, and the number of terms is n + 1. The quantities j), q, t, u, v are either positive or nega- tive, integral or fractional ; and the coefficient of x 11 is unity. 481. In reducing an equation to the general form, all the terms must be transposed to the first member, and arranged according to the powers of x. If as" has a coefficient, it may be removed by dividing the equation by that coefficient. 370 ALGEBRA. 482. A Root of an equation is any real or imaginary ex- pression, which, being substituted for its unknown quantity, satisfies the equation, or makes the first member equal to (Art. 166). We assume that every equation has at least one root. 483. An equation of the third degree containing only one unknown quantity, or one in which the cube is the highest power of the unknown quantity, is usually called a cubic equa- tion. 484. An equation of the fourth degree containing only one unknown quantity is usually called a biquadratic equation. DIVISIBILITY OF EQUATIONS. 485. If a is a root of an equation in the form x n +qi x n ~ l + q x n ~ 2 + + tx 2 + ux + V — 0, then the first member is divisible by x — a. It is evident that the division of the first member by x — a maybe carried on until x disappears from the remainder. Let Q represent the quotient, and R the remainder, which is inde- pendent of x ; then the given equation may be made to take the form . (x - a) Q + B = 0. But if x = a, then (x — a) Q = 0, and, consequently, R = 0; that is, x — a is a factor of the first member of the given equa- tion, as it is contained in it without a remainder. 486. Conversely, if the first member of am. equation in the form * x n -\-px n ~ 1 -\- qx n ~ 2 + + t x 2 + ux + v = is divisible by x — a, then a is a root of the equation. GENERAL THEORY OF EQUATIONS. 371 For, if the first member of the given equation is divisible by x — a, then tbe equation may be made to take the form (x — a) Q = ; and it follows from Art. 330 that a is a root of this equation. EXAMPLES. By the method of Art. 486, 1. Prove that 3 is a root of the equation x 3 — 6 x 2 + 11 x — 6 = 0. 2. Prove that — 1 is a root of the equation x 3 + 1 = 0. 3. Prove that 1 is a root of the equation x 3 + x 2 — 17 a; + 15 = 0. 4. Prove that — 2 is a root of the equation a; 4 — 3 x 2 + 4 x + 4 = 0. 5. Prove that 4 is not a root of the equation x* — 5 x 3 + 5 x 2 + 5 x - 6 = 0. NUMBER OF ROOTS. 487. Every equation of the nth degree, containing but one unknown quantity, has n roots, and no more. Let a be a root of the equation x n +2) x n ~ x + qx n ~ 2 + + tx 2 + ux + v — 0; then, by Art. 485, the first member is divisible by x — a, and the equation may be made to take the form (x — a) (V 1 - 1 + 2h x n ~ 2 + + «j x + i\) = 0. The equation may be satisfied by making either factor of the first member equal to (Art. 330) ; hence, x • — a = and x»~ 1 +p 1 x n - i + +u l x + v 1 = 0. (1) 372 ALGEBRA. But equation (1) must have some root, as b, and may be placed under the form (x — b) (x 11 - 2 +jh x n ~ 3 + + v,x + v,)=0', which is satisfied by placing either factor of the first member equal to ; and so on. Since each of the factors x — a, x — b, etc., contains only the first power of x, it is evident that the original equation can be separated into as many such binomial factors as there are units in the exponent of the highest power of the unknown quantity, and no more ; that is, into n factors, or (x — a) (x — b) (x — c) (x — T) = 0. Hence, by Art. 330, the equation has the n roots a, b, c, /. Moreover, if the equation had another root, as r, then it must contain another factor x — r, which is impossible. 488. It should be observed that the n binomial factors of which the general equation of the nth degree is composed, are not necessarily unequal j hence, two or more of the roots of an equation may be equal. Thus, the equation x * _ g x 2 + 12 x _ 8 = may be factored so as to take the form (x-2)(x-2) (.r-2)=0, or (*-2) 3 = 0; and hence the three roots are 2, 2, and 2. 489. It will be readily seen that any equation, one of whose mots is known, may be depressed to another of the next lower degree, which shall contain the remaining roots. Hence, if all the roots of an equation are known excepl two, those may be obtained from the depressed equation, hy the rules for quadratics. 1. One root of the equation x s + 2 x 1 — 23 x — 60 = is — 3 ; what are the others ? GENERAL THEORY OF EQUATIONS. 373 Dividing x 3 + 2 x~ — 23 x — 60 by x + 3, the given equation may be put in the form O + 3)O 2 -x-20)=0. Thus,- the depressed equation is x 2 — x — 20 = 0. Solving this by the rules for quadratics, we obtain x = 5 or — 4 ; which are the remaining roots. EXAMPLES. 2. One root of the equation x z — 19 x + 30 = is 2 ; what are the others ? 3. Eequired the three roots of the equation X s = a 3 , or x 3 — a 3 — 0. 4. One root of the equation x 3 + x 2 — 16 x + 20 = is — 5 ; reqxiired the remaining roots. 5. Two roots of the equation x A — 3 x 3 — 14 x" + 48 x — 32 = are 1 and 2 ; required the remaining roots. 6. One root of the equation x 4 — 7 x 3 + 3 x + 3 = is 1 ; what equation contains the remaining roots ? 7. One root of the equation 6 x 3 — x~ — 32 x + 20 = is 2 ; what are the others ? 8. Two roots of the equation 20 x* - 169 x 3 + 192 x~ + 97 x — 140 = are 1 and 7 : what are the others ? 5 FORMATION OF EQUATIONS. 490. An equation having any given roots may be formed by subtracting each root from the unknown quantity, and pla- cing the product of these binomial factors equal to 0. For it is evident, from principles already established, that an equation having the n roots a, b, c, I may be written in the form (x — a) (x — b) (x — c) (x — I) = 0. 374 ALGEBRA. After performing the multiplication indicated, the equation will assume the form . x n +2?x n ~ 1 + qx n ~ 2 + + tx 2 + ux + v = 0. (Compare Art. 329.) 1. Form the equation whose roots are 1, 2, and — 4. Result, (x — 1) (cc — 2) (cc + 4) = or, x 3 + x 2 — 10 x + 8 = 0. EXAMPLES. Form the equations whose roots are : 2. - 1, - 3, and - 5. 6. 1, 2, 3, and 4. 3. 5, - 2, and - 3. 7. 4, 4, and 5. 4. 1, - , and - . 8. 0, - 1, 3, and 4. 5. ± 1 and ±2. 9. - 5 ? , - 2, and ? 4 3 COMPOSITION OF COEFFICIENTS. 491. The coefficient of the second term of an equation of the nth degree in its general form is the sum of all the roots with their signs changed; that of the third term is the sum of their products, taken two and two ; that of the fourth term is the sum of their 'products, taken three and three with their .signs changed, etc. ; and the last term is the product of all the roots with their signs changed. For, resuming the equation (x '— a) (x — b)(x — c) (x — k) (x — l)=0, if we perform the multiplication indicated, we obtain (x — a) (x — b) = x~ — (a + b) x + a b, (x—a)(x — b) (x—c) = x 3 —(a + b+c)x+(ab + ac+bc)x—abc, GENERAL THEORY OF EQUATIONS. 375 and so on. When n factors have heen multiplied, the coeffi- cients of the general equation become jp = — a — b —c— — k — I q = ab + ac-\-bc-\- + kl r = — a b c — ab d — a c d — — ikl v = ±a b c kl which corresponds with the enunciation of the proposition ; the upper sign of the value of v being taken when n is even, and the lower sign when n is odd. 492. If ^=0, that is, if the second term of an equation be wanting, the sum of the roots will be 0. If v = 0, that is, if the absolute term of an equation be want- ing, at least one root must be 0. 493. Every rational root of an equation is a divisor of the last term. 494. When all the roots of an equation but two are known, the coefficient of the second term of the depressed equation (Art. 489) can be found by subtracting the sum of the known roots, with their signs changed, from the coefficient of the second term of the original equation. The absolute term of the depressed equation can be found by dividing the absolute term of the original equation by the product of the known roots with their signs changed. EXAMPLES. Find the sum and product of the roots in the following : 1. a; 3 -7a;+6 = 0. 2. 2 x*- 5 x s - 17 x 2 + 14 x + 24 = 0. In the following example obtain the depressed equation by the method of Art. 494 : 3. Two roots of the equation x 4 — 5 x s — 2 x 2 + 12 x + 8 = are 2 and — 1 ; what are the others ? 376 ALGEBRA. FRACTIONAL ROOTS. 495. An equation whose coefficients are all integral, the coefficient of the first term being unity, cannot have a rational fraction as a root. If possible, let -, a rational fraction in its lowest terms, be a root of the equation x n + px n ~ l + qx n ~ 2 + + tx 2 + UX + v = 0, where p, q, , t, u, v are integral. Then a\ n faV' 1 fa\ n - 2 ( a\ 2 fa Multiplying through by lf~ l , and transposing, - n = — ( 2 ,a n - 1 +qa n - 2 b + + ta 2 b n ~ s + ua u n ~ 2 + vb n ~ l ). Now, as - is in its lowest terms, a and b can have no com- mon divisor ; therefore a n and b can have no common divisor ; a" hence — is in its lowest terms. Thus, we have a fraction in o its lowest terms equal to an entire quantity, which is impossi- ble. Therefore no root of the equation can be a rational fraction. Note. The equation may have an irrational fraction as a root, such as ' for example. Such a root, whose value can only be expressed 4 approximately by a decimal fraction, is called incommensurable. IMAGINARY ROOTS. 496. If the coefficients of an equation be real quantities, imaginary roots enter it by pairs, if at all. Suppose a + b ^ — 1 to be a root of the equation X n + p X" ~ l + q X n ~ 2 + + t X 2 + U X + V = 0. GENERAL THEORY OF EQUATIONS. 377 Substituting a + b y/— 1 for x, and developing each expres- sion by the Binomial Theorem, all the odd terms of each series will contain either powers of a, or even powers of b y/ — 1, and are therefore real ; while all the even terms contain the odd powers of b y — 1, and are therefore imaginary. Representing the sum of all the real quantities by P, and the sum of all the imaginary quantities by Q y — 1, we have P+ £y/-l=0. This equation can be true only when both P and Q equal 0. If we now substitute a — b y/ — 1 for x, we find that the series differ from the former only in having their even or imaginary terms negative. Hence, we obtain as the first member P-Q^-l, which must be equal to 0, for we have already shown that both P and Q equal 0. Thus, a — b y/— 1 satisfies the equation. Similarly, we may show that, if b y/ — 1 is a root of the equa- tion, then will — b y/— 1 also be a root of the equation. 497. The product of a pair of imaginary quantities is always positive. Thus, (a + 6y/-l) (a-b^-l) = a 2 + b and y/^l) (- b y/^I) = b\ TRANSFORMATION OF EQUATIONS. 498. To transform an equation into another which shall have the same roots with contrary signs. Let the given equation be x n + p x n ~ 1 + q x n ~ 2 + + t x 2 + u x + v = 0. 378 ALGEBRA. Put x = — y; then whatever value x may have, y will have the same value with its sign changed. The equation now becomes {-y) n +p(-y) n - l + q(-y) n - 2 + + t(-y) 2 + u{- y ) + v = o. If 11 is even, the first term is positive, second term nega- tive, and so on ; and the equation may be written yn_ p yU-\ + (1 yn~1_ + f y'2 _ ^ y _J_ y _ Q Qj If n is odd, the first term is negative, second term positive, and so on ; hence, changing all signs, we write the equation y" —v y n ~ 1 + 1 y n ~ 2 — — ^ y 1 + u y~ v — o. (2) From (1) and (2) it is evident that to effect the desired transformation we have simply to change the signs of the alternate terms, beginning with the second. Note. The preceding rule assumes that the given equation is complete (Art. 300) ; if it be incomplete, any missing term must be put in with zero as a coefficient. 1. Transform the equation x 3 — 7x + 6 = into another which shall have the same roots with contrary signs. We may write the equation x 3 + . ar — 7 x + 6 = 0. Applying the rule, x s _ o . x 2 — 7 x — 6 = 0, or x z — 7 x — 6 = 0, Ans. EXAMPLES. Transform the following equations into others which shall have the same roots with contrary signs: 2. x * - 2 x s + x - 1.32 = 0. 3. x 5 -3x 2 + 8 = 0. 499. To transform an equation into another whose roots shall be some multiple of those of the first. Let the given equation he x n +px n - 1 + qx n ~ 2 + + tx' 2 + ux + v = 0. GENERAL THEORY OF EQUATIONS. 379 v Put x = — ; then whatever value x may have, y will have m a value m times as great. The equation now becomes (i)r + jjLY-\Ji\"-\ +,(£)*+.(»)+.=* Multiplying through by ra n , we have y n +pmy n ~ 1 +qm 2 y n ~ 2 + + tm n ~ 2 y 2 + um n ~ 1 y + vm" = 0. Hence, to effect the desired transformation, multiply the second term by the given factor, the third term by its square, and so on. Similarl}'-, we may transform an equation into one whose roots shall be those of the first divided by some quantity. 1. Transform the equation x 3 — 7 x — 6 = into another whose roots shall be 4 times as great. The equation may be written, x s -f- . x 2 — 7 x — 6 = 0. Then, by the rule, x 3 - 4 2 . 7 x - 4 3 . 6 = 0, or x 3 - 112 x - 384 = 0, Ans. EXAMPLES. 2. Transform the equation x z — 2 x 2 + 5 = into another whose roots shall be 5 times as great. 3 x s 3. Transform the equation x i -| — 27 = into another whose roots shall be one third as great. 500. Ta transform an equation containing fractional coefficients into another whose coefficients are integral, that of the first term being unity. If in Art. 499 we assume m equal to the least common multiple of the denominators, it will always remove them ; but often a smaller number can be found which will produce the same result. 380 ALGEBRA. 1. Transform the equation x s — — — — + — — = into o So lUo another whose coefficients shall he integral. The least common multiple of the denominators is 108 ; so that one solution would he, by Art. 499, rf_M8. |-.M8-:| + 108^ = 0. An easier way, however, is as follows ; the denominators may be written 3, 3 2 X 2' 2 , and 3 3 X 2 2 , so that the multiplier 3 x 2 or 6 will remove them. Hence, by Art. 499, we have x a_ 6 * a __ 6 3 * +6 s * =0 or x *-2x*-x + 2 = 0, ob 106 whose roots are 6 times as great as those of the given equation. EXAMPLES. Transform the following equations into others whose coef- ficients shall be integral : **'+T~ 7 4= - *.- , + re-g-»= a av-^ + fwb. 5.^-5^-^ + 5 = 0. no 4 I 501. To transform an equation into another ivhose roots shall be the reciprocals of those of the first. Let the given equation be x n + p x n ~ 1 + q x n ~ - + + t x 2 + u x + v = 0. Put x = - ; then whatever value x may have, y will be its v reciprocal. The equation now becomes 1 p q t u n y y y y y GENERAL THEORY OF EQUATIONS. 381 Multiplying through by y", and reversing the order, vy n + uy n ~ 1 + ty"- 2 + + qy 2 +py + 1 = 0. Dividing through by v, u , t a „ 7} 1 n + _ y «-i + - gf-» + + Lf + L l/+ ± =0 . v « w «; «; Hence, to effect the transformation, write the coefficients in reverse order, and then divide by the coefficient of the first term. EXAMPLES. Transform the following equations into others whose roots shall he the reciprocals of those of the first : 1. a; 3_6x 2 +ll iC -6 = 0. 3. x 3 -9x 2 + — -i = 0. 7 49 2. x i ~x 3 -3x 2 + x + 2 = 0. 4. x 3 -4x 2 + 9 = 0. 502. To transform an equation into another whose roots shall differ from those of the first by a given quantity. Let the given equation be s"+ju""' + qx n ~ 2 + + tx* + ux + v = 0. (1) Put x = y + r, and we have (ff + r)*+p(t,. + r)— 1 + + u (y + r ) + v=0. (2) Developing (y + r) n , (y + r)"-\ , by the Binomial The- orem, and collecting terms containing like powers of y, we have an equation of the form yn +pi yn-l + ^ y «-1 + + ^ y 2 + ^ y + ^ = Q ( o) As y=x — r, the roots of (3) are evidently less by r than those of (1). By putting x = y — r, we shall obtain in the same way an equation whose roots are greater by r than those of (1). 503. If n is small, the operation indicated in Art. 502 may be effected with little trouble; but for equations of a higher degree a less tedious method is better. 382 ALGEBRA. If in (3) we put y = x — r, we shall have (x—rf +2h(x — r) n - 1 + +u 1 (x — r) + v 1 = 0, (4) which is, of course, identical with (1), and must reduce to (1) when developed. If we divide (4) by x — r, we obtain (x - r)" ~ J +jh - r) n - 2 + q i {x-r)"-"+ +u, (5) as a quotient, with a remainder of v v Dividing (5) by x — r, we obtain a remainder of u x ; and so on, until we obtain all the coefficients of (3) as remainders. Hence, to effect the desired transformation, Divide the given equation by x — r or x + r> according as the roots of the transformed equation are to be less or greater than, those of the first by r, and the remainder will be the absolute term of the transformed equation. Divide the quo- tient just found by the same divisor, and the remainder will be the coefficient of the last term but one of the transformed equation ; and so on. 504. 1. Transform the equation x 3 + 3 x 2 — 4^ + 1 = into one whose roots shall be greater by 1. Using the method of Art. 502, put x = y — 1. Then, (y-iy + 3(y~iy-4(y-l) + l = 0, or, t/ 3 - 3 y 2 + 3 7/ - 1 + 3 / - 6 y + 3 - 4 y + 4 + 1 = 0, or, if — 7 y + 7 = 0, Ans. EXAMPLES. 2. Transform the equation x 3 — x — 6 = into one whose roots shall be less by 8. 3. Transform the equation x* -f 6 x 3 — x 1 — 5 x — 1 = into one whose roots shall be greater by 3. 505. To transform a complete equation into one tchos* second term shall be wanting. GENERAL THEORY OF EQUATIONS. 383 The coefficient of y" -1 in (2), Art. 502, is n r + p. Hence, in (3), p y = n r + p. To make ^ = 0, it is only necessary to make nr + p =0, or r = — — ; hence, to effect the desired n V transformation, put x = y — - ; that is, put x equal to y } minus the coefficient of the second term of the given equation divided by the degree of the equation. 1. Transform the equation x 3 — G x~ -j- 9 x — G = into another whose second term shall be wanting. Here p = — 6, n = 3 ; then, put x = y — — — = y + 2. Result, (y + 2) 3 - 6 (y + 2) 2 + 9 (y + 2) - 6 = 0, or, tf + 6y* + 12y + 8-6y*-24:y-24:+9y +18-6 = 0, or, y 3 — 3 y — 4 = 0, Ans. EXAMPLES. Transform the following equations into others whose second terms shall be wanting : 2. x 2 -px + q = 0. 4. x 3 + 6x 2 -3x + 4: = 0. 3. x s + x 2 + 4 = 0. 5. x 4 - 4 x 3 - 5 x — 1 = 0. DESCARTES' RULE OF SIGNS. 506. A Permanence of sign occurs when two successive terms of a series have the same sign. A Variation of sign occurs when two successive terms of a series have contrary signs. DESCARTES' RULE. 507. A complete equation cannot have a greater number of positive roots than it has variations of sign, nor a greater number of negative roots than it has permanences of sign. 384 ALGEBRA. Let any complete equation have the following signs : + + + - + - + in which there are three permanences and five variations. If we introduce a new positive root a, we multiply this by x — a (Art. 490). Writing only the signs which occur in the operation, we have + + + - + — 1 + — + + + - + - + — + - + — + + + ± ± — + — + — ± + 123456789 10 a double sign being placed wherever the sign of a term is ambiguous. However the double signs are taken, there must be at least one variation between 1 and 4, and one between 8 and 10, and there are evidently four between 4 and 8 ; or in all there are at least six variations in the result. As in the original equation there were five variations, the introduction of a positive root has caused at least one additional variation ; and as this is true of any positive root, there must be at least as many variations of sign as there are positive roots. Similarly, hy introducing the factor x + a, we may show that there are at least as many permanences of sign as there are negative roots. , If the equation is incomplete, any missing term must be supplied with ± as its coefficient before applying Descartes' Rule. 508. In any complete equation, the sum of the number of permanences and variations is equal to the number of terms less one, or is equal to the degree of the equation (Art. 480). Hence, when the roots are all real, the number of positive roots is equal to the number of variations, and the number of negative roots is equal to the number of permanences (Art. 4S7). GENERAL THEORY OF EQUATIONS. 385 A complete equation whose terms are all positive can have no positive root ; and one whose terms are alternately positive and negative can have no negative root. 509. In an incomplete equation, imaginary roots may sometimes he discovered hy means of the douhle sign of in the missing terms. Thus, in the equation x 3 + x 2 ± x + 4 = if we take the upper sign, there is no variation, and conse- quently no positive root; if we take the lower sign, there is hut one permanence, and hence but one negative root. There- fore, as the equation has three roots (Art. 487), two of them must he imaginary. In general, whenever the term which precedes a missing term has the same sign as that which follows, the equation must have imaginary roots ; where it has the opposite sign, the equation may or may not have imaginary roots, hut Descartes' Hule does not detect them. If two or more suc- cessive terms of an equation he wanting, there must be imagi- nary roots. Note. In all applications of Descartes' Rule, the equation must con- tain a term independent of x, that is, no root must be equal to zero (Art. 330) ; for a zero root cannot be considered as either positive or negative. EXAMPLES. 510. The roots of the following equations being all real, determine their si cms : *&' 1. x 8 -3x-2 = 0. 3. a; 8 -7 a; 2 + 36 = 0. 2. .x 3 -10a; + 3 = 0. 4. x i -2x 3 -13x 2 + 38*-24 = 0. 5. What are the signs of the roots of the equation x 3 + x 1 -4 = 0? DERIVED POLYNOMIALS. 511. If we take the polynomial a x n + b x n ~ 1 + c x n ~ 2 + 386 ALGEBRA. and multiply each term by the exponent of a; in that term, and then diminish the exponent by 1, the result n a x" ~ l + (n — 1) b x H ~ 2 + Qi — 2) c x n ~ 3 + is called the first derived polynomial or first derivative of the given polynomial. The second derived polynomial or second derivative is the first derived polynomial of the first derivative ; and so on. The given polynomial is sometimes called the primitwe poly- nomial. A derived equation is one whose first member is a deriva- tive of the first member of another. 1. Find the successive derivatives of x* + 5 x 1 + 3 x + 9. Result : First, 3 x 2 + 10 x + 3. Second, 6 x + 10. Third, 6. Fourth, 0. EXAMPLES. Find the successive derivatives of the following : 2. a; 3 -5r + 6x-2. 4. a x* -bx % + ex -Sd. 3. 2cc 2 -ic-7. 5. 7cc 4 -13a; 2 +8x-l. EQUAL ROOTS. 512. Let the roots of the equation x n +px n -' 1 + qx n ~ 2 + + tx 2 + ux + v = (1) be a, b, c, Then (Art. 490), we have x n +2? x n ~ 1 + q x n ~ 2 + = (x — a) (x — b) (x — c) Putting x + y in place of x, (x + y) n + p (x + y)"- 1 + ... = (y + x -a) (y + x-b) ... (2) By Art. 399, the coefficient of y in the first member is nx n ~ 1 +p (n — I) x n ~ 2 + q (n — 2) x n ~* + (3) GENERAL THEORY OF EQUATIONS. 387 which, we observe, is the first derivative of (1) ; and, as in Art. 491, regarding x — a, x — b, as single terms, the coefficient of y in the second member is (x — b) (x — c) (x — d) to n — 1 factors "1 + (x — a) (x — c) (x — d) to n — 1 factors ! + (x — a) (x — b) (x — d) to ii — 1 factors | ^ ' + ...... j As (2) is identical, by Art. 413 these coefficients are equal. Now if b = a, that is, if equation (1) has two roots equal to a, every term of (4) will be divisible by x — a, hence (3) will be divisible by the same factor ; therefore (Art. 486) the first derived equation of (1) will have one root equal to a. Sim- ilarly, if c = b = a, that is, if (1) has three roots equal to a, (3) will have two roots equal to a ; and so on. Or, in general, If an equation has n roots equal to a, its first derived equa- tion xv ill have n — 1 roots equal to a. 513. From the principle demonstrated in Art. 512, it is evident that to determine the existence of equal roots in an equation we must Find the greatest common divisor of the first member and its first derivative. If there is no common divisor there can be no equal roots. If there is a greatest common divisor, by placing it equal to zero and solving the resulting equation we shall obtain the required roots. The number of times that each root is found in the given equation is one more than the number of times it is found in the equation formed from the greatest common divisor. If the first member of the given equation be divided by the greatest common divisor, the depressed equation will contain the remaining roots of the original equation. 1. Find the roots of the equation x* - 14 x s + 61 x~ - 84 x + 36 = 0. Here the first derivative is 4 x s — 42 x % + 122 x — 84 ; the greatest common divisor of this and the given first member 388 ALGEBRA. i s x 1 — 7 x + 6. Placing x 2 — 7 x + 6 == 0, we have, by the rules of quadratics, or by factoring, x = 1 or 6. Therefore the roots of the given equation are 1, 1, 6, and 6. EXAMPLES. Find all the roots of the following : 2. x s-Sx-+13x-6 = 0. 4. .T 4 -6a; 2 -8a;-3 = 0. 3. x 3 -7x 2 +16x-12 = Q. 5. cc 4 -24x 2 + 64z-48 = 0. 514. When the equation formed from the greatest com- mon divisor is of too high a degree to be conveniently solved, we may in certain cases compare it with its own derived equation, and thus obtain a common divisor of a lower degree. Of course this can only be done when the equation formed from the greatest common divisor has equal roots. For example, required all the roots of x b - 13 x* + 67 x 3 - 171 x- + 216 x - 108 = 0. (1) Here the first derivative is 5 x* — 52 x 3 + 201 ,x 2 — 342 x + 216 ; the greatest common divisor of this and the given first member is x 3 — 8 x 2 + 21 x — 18. We have then to solve the equation x 3 -8x 2 +21z-18 = 0. (2) The first derivative of (2) is 3 x 2 — 16 x + 21 ; the greatest common divisor of this and x 3 — 8 x 2 + 21 x — 18 is x — 3. Solving x — 3 = 0, we have x = 3; hence two of the roots of (2) are equal to 3. Dividing the first member of (2) by (x — 3) 2 or by x 2 — 6 x + 9, the depressed equation is x — 2 = 0, whence x = 2. Thus the three roots of (2) are 3, 3, and 2. Hence, the five roots of (1) are 3, 3, 3, 2, and 2. 515. If an equation has two roots equal in magnitude, but opposite in sign, by changing the signs of the alternate terms beginning with the second we shall obtain an equation with these same two roots (Art. 498) ; then evidently the greatest GENERAL THEORY OF EQUATIONS. 389 common divisor of the two first members placed equal to zero will determine the roots. For example, required all the roots of x* + 3 x 3 - 13 x--21 x + 36 = 0. (1) Changing the signs of the alternate terms, we have x 4 - 3 x 3 - 13 x°- + 27 x + 36, the greatest common divisor of which and the given first member is x 2 — 9 ; solving x 2 — 9 = 0, we have x = 3 or — 3. thus giving two of the roots of (1). Dividing the first mem- ber of (1) by x 2 — 9, we have for the depressed equation x- + 3 x - 4 = 0, whence x = 1 or — 4. Thus the roots of (1) are 3, — 3, 1, or —4. LIMITS OF THE ROOTS OF AN EQUATION. 516. A polynomial of the form x n + p x n ~ * + q x n ~ 2 + J r tx 2 + ux+v which we shall represent by X, may also be expressed thus (Art. 490) : (x — a) (x — b) (x — c) (x — I) Y in which a, b, c, I are the real, unequal roots of the equa- tion X= 0, in the order of their magnitude, a being algebrai- cally the smallest; and Y the product of all the factors con- taining imaginary roots, which must always be positive, and cannot affect the sign of X, for each pair of imaginary roots (Art. 497) produces a positive factor. Suppose x to commence at any value less than a, and to assume in succession all possible values up to some quantity greater than I. When x is less than a, each of the factors x — a, x — b, is negative, and therefore X is either posi- tive or negative, according as the degree is even or odd. 390 ALGEBRA. When x == a, X = 0. When x is greater than a, and less than b, x — a becomes positive, and the sign of X changes. Also, when the value of x is made equal to b, and then greater, X first becomes and then changes sign; and so on, for each real root. When x has any value greater, than I, X must be positive ; for all its factors are positive. 517. If two numbers, when substituted for the unknown quantity in an equation, give results having a different sign, at least one root lies between those numbers. It is evident, from Art. 516, that if X has a different sign for two values of x, some odd number of roots lies between them. When the numbers substituted differ by unity, it is evident that the integral part of the root is known. EXAMPLES. 1. What is the first figure of a root of the equation x 3 + 3 x- Here, if x = 2, the first member becomes — 2 ; and if x = 3, the first member becomes 25 ; therefore at least one root lies between 2 and 3. Hence 2 is the first figure of a root. 2. Find the integral parts of all the roots of the equation x 3-Gx 2 +3x+9 = 0. 3. Find the first figure of a root of the equation x 3 —2x - 50 = 0. 4. Find the first figure of a root of the equation x* — 2 ar 8 + 3x 2 -x-5 = 0. 5. Find the integral part of a root of the equation 2 x 4 + x : -7x 2 -llx-4, = 0. 518. To find the superior limit of the positive roots of an equation. Let the equation be X= x n + p x"- 1 + q .r"- 2 + + tx- + ux + v = 0. (1) a GENERAL THEORY OF EQUATIONS. 301 Let r be the numerical value of the greatest negative coefficient, and x n ~ s the highest power of x which has a nega- tive coefficient. Then the first s terms have positive coef- ficients. Now Xwill be positive when x is positive, provided x 11 — r x n ~ * — ) , / _,_1 — — r x- — r x — r (2) is positive ; for, since r is the numerically greatest negative coefficient, and all terms up to the (s + l)th are positive, Xis equal to (2) plus a, positive quantity. We may write (2) x n — r (x n - s + x n - s - 1 + + x °- + x+l), or (Art. 120), x*-r - — — ~ . (3) x — 1 Then AT will be positive when (3) is positive. But if x is greater than unity, (3) is evidently greater than x n — s + 1 X-l Therefore X will be positive when this is positive ; or, when (x — 1) x 11 — rx n - s + 1 is j>ositive ; or, when (x — 1) x s ~ x — r is positive. But (x— 1) X s-1 — r is greater than (x — 1) (x — l) s_1 — r or (x — 1)' — r ; therefore X will be positive when (x — l) s — r is positive or equal to zero ; or, when (x — l) s = r or > r ; or, when x — l — \Jr or > \/r; or, when x = l-\-\jr or >1 + </r. That is, when x = 1 + \J r or any greater value, X is posi- tive, which is impossible, as it must equal zero. Hence x must be less than l + tyr; or, 1 + $r is the superior limit of the positive roots. 519. To find the inferior limit of the negative roots of an equation. By changing the signs of the alternate terms beginning with the second, we shall obtain an equation having the same roots with contrary signs (Art. 408). 392 ALGEBRA. Then evidently the superior limit of the positive roots of the transformed equation, obtained as in Art. 518, will by a change of sign become the inferior limit of the negative roots of the given equation. Note. In applying the principles of the preceding articles to determine the limits of the roots of an equation, the absolute term must be taken as the coefficient of x'K 520. 1. Find the superior limit of the positive roots of x 4 + 4 x 3 - 19 x- - 46 x + 120 = 0. Here, r = 46, and n — s = 2 ; or, as n = 4, s — 2. Then by Art. 518, the required limit is 1 + y/46, or 8 in whole num- bers. 2. Find the inferior limit of the negative roots of a- 3 -a; 2 -14.x +24 = 0. (1) Changing the signs of the alternate terms beginning with the second, we have x 3 + x 2 -Ux-2A = 0. (2) Here r = 24, and n — s = 1, or s = 2. Then the superior limit of the positive roots of (2) is 1 + \/24 ; therefore the inferior limit of the negative roots of (1) is — (1 + v/24). EXAMPLES. Find the superior limits of the positive roots of the follow- ing: 3. x * + 2x 3 -13.r' 1 -Ux + 24:=Q. 4. a: 4 -15.T 2 +10,r+24=0. Find the inferior limits of the negative roots of the fellow- s' mg 5. x*-2x°--5x + (j = 0. 6. x x -5x 5 +ox- + 5x + 6 = 0. STURM'S THEOREM. 521. To determine the number and situation of the real roots of an equation. GENERAL THEORY OF EQUATIONS. 393 A perfect solution of this difficult problem was first obtained by Sturm, in 1829. As the theorem determines the number of real roots, the number of imaginary roots also becomes known (Art. 487). 522. Let X denote the first member of X n +21X n - 1 + qx n ~ 2 + + tX 2 + UX+ V = 0, from which the equal roots have been removed (Art. 512). Let X x denote the first derivative of X (Art. 511). Divide AT by X 1} and we shall obtain a quotient Q l} with a remainder of a lower degree than X v Denote this remainder, with its signs changed, by X 2 , divide X x by X 2 , and so on ; the operation being the same as in finding the greatest com- mon divisor, except that the signs of every remainder must be changed, while no other change of signs is admissible. As the equation X — has been freed from equal roots, there can be no common divisor of Zand J 1; and the last remainder, X„ , will be independent of x. The successive operations may be represented by the fol- lowing equations : X = X x Q v - X 2 (1) X I = X 2 Q 2 -X 3 (2) X^X,Q,-X 4 (3) X n _ 2 — -<*« - 1 V« - 1 X n The expressions X, X 1} X 2 , X H are called Sturm's Functions. STURM'S THEOREM. 523. If any tiro numbers, a and b, be substituted/or x in Sturm's Functions, and the signs noted, the difference between the number of variations in the first case and that in the second is equal to the number of real roots of the given equa- tion lying between a and b. 394 ALGEBRA. The demonstration of Sturm's Theorem depends upon the following principles : (A). Two consecutive functions cannot both become for the same value of x. For, if A\ = and X = 0, then by (2), Art. 522, X 3 = ; and if X 2 = and X 3 = 0, by (3), X 4 = 0; and so on, till X n = 0. But as X n is independent of x, it cannot become for any value of x. Hence no two consecutive functions can become zero for the same value of x. (B). If any function, except X and X n , becomes for a ■particular value of x, the two adjacent functions must have opjjosite signs. For, if X, = 0, we have by (2), Art. 522, X 1 = — X s ; that is, X 1 and X s must have opposite signs, for by (A) neither can be equal to zero. (C). When any function, except X and X n , changes its sign for different values of x, the number of variations is not affected. No change of sign can take place in any one of Sturm's Functions except when x passes through a value which re- duces that function to zero. Now, let c be a root of the equation X 2 = 0; d and e quan- tities respectively a little less and a little greater than c, so taken that no root of X x = or of X s = is comprised be- tween them. Then, as x changes from d to e, no change of sign takes place in A^ or X :i , while X 2 reduces to zero and may change sign. And as by (B), when X 2 = 0, X l and X z have opposite signs, the only effect of a change in the sign of X 2 is that what was originally a permanence and a variation is now a variation and a permanence ; that is, the permanence and variation exchange places. Hence a change in the sign of X., does not affect the number of variations. As X n is independent of x, it can never change sign for any value of x. Therefore a change in the number of variations GENERAL THEORY OF EQUATIONS. 395 can be caused only by a change in the sign of the given function X. (D). When the function X changes its sign for successive increasing values of x, the number of variations is diminished by one. Let m be a root of the equation X = ; m — y and m + y quantities respectively a little less and a little greater than 7>i, so taken that no root of X 1 = Q is comprised between them. Then, as x changes from m — y to m + y, no change of sign takes place in X r , while X reduces to zero and changes sign. Putting in + y in place of x in X, we have O + y) n + P (m + y) n ~ l + + u (m + y) + v. B/eveloping the terms by the Binomial Theorem, and col- lecting terms containing like powers of y, we have m n + p m n ~ 1 + + um + v + y \_n m" _ l + p (n — 1) ni n ~~ 2 + + u\ + terms containing y 2 , y s , y n . Representing the coefficient of y, which we observe is the value of X 1 when x is put equal to m, by A ; the coefficient of y 1 by B; and so on, we have m n +pm n - 1 + + um + v + Ay + By-+ + Ky n . (1) But as x = m reduces X to 0, we have m n + p m n ~ 1 + + u m + v = 0. Hence (1) may be written Ay + Bf-+ + Ky\ (2) Now y may be taken so small that the sign of (2) will be the same as the sign of its first term. That is, when a? is a little greater than m, the sign of X is the same as the sign of X l . Similarly, by substituting in — y for x in X, we shall arrive at the expression -Ay+By 2 -Cy 3 + , 396 ALGEBRA. where as before y may be taken so small that the sign of the whole expression will be the same as that of its first term. That is. when a; is a little less than m, the sign of X is the reverse of the sign of X x . Thus we see that as x changes from m — y to m + y, the signs of X and A^ are different before x equals m, and alike afterwards. Hence, when A changes its sign a variation is changed into a permanence, or the number of variations is diminished by one. We may now prove Sturm's Theorem ; for as x changes from a to b, supposing a less than b, a variation is changed to a permanence each time that X reduces to and changes sign, and only then, for no change of sign in any of the other functions can affect the number of variations. And as X reduces to zero only when x is equal to some root of the equation X — 0, it follows that the number of variations lost in passing from a to l> is equal to the number of real roots of the equation X = comprised between a and b. 524. When — oo and + go are substituted for x, or when the superior limit of the positive roots and the inferior limit of the negative roots are substituted for x, the whole number of real roots of the equation A^ = becomes known. The substitution of — go and will give the whole number of negative roots, and the substitution of + go and will give the whole number of positive roots. If the roots are all real, Descartes' Kule (Art. 507) will effect the same object. The substitution of various numbers for x will show be- tween what numbers the roots lie, or fix the limits of the roots. 525. A and A, must change signs alternately, as they are always unlike in sign just before X changes sign (Art. f>L\' > > i (D)). Hence, when the roots of X = and of Aq = are all real, each root of Aq = must be intermediate in value be- tween two roots of X= 0. For this reason the first derived equation is often called the limiting or separating equa- tion. GENERAL THEORY OF EQUATIONS. 397 526. In the process of finding X 2 , X 3} etc., any positive numerical factors may be omitted or introduced at pleasure, as the sign of the result is not affected thereby. In this way fractions may be avoided. In substituting — go and + go, the first term of each func- tion determines the sign, for in any expression, as ax n + bx n ~ 1 + + /,•, where x may be made as great as we please, it may be taken so great that the sign of the whole expression will be the same as that of its first term. 527. 1. Determine the number and situation of the real roots of the equation x 3 — 4 x 2 — x + 4 = 0. Here, the first derivative, X 1 = 3 x 2 — 8 x — 1. Multiply- ing x 3 — 4 x' 1 — x + 4 by 3 so as to make its first term divisi- ble by 3 x 2 , 3x 2 -8x-l)3x 3 -12x 2 - 3 a: + 12 (a; 3 a; 3 - Sx 2 x - 4r- 2 a; + 12 3 - 6 a; 2 - 3 x + 18 (-2 - 6.r- + 16a- + 2 19 a: + 16 .-. X = 19 x - 16. 3 x 2 - 8 x - 1 19 19 x - 16 ) 57 x 2 - 152 x - 19(3x 57 x 2 — 48 x - 104 x - 19 19 1976 a;- 361 (-104 1976 x + 1664 -2025 .-. X = 2025. 398 ALGEBRA. Thus we have, X = x z — 4 x 2 — x + 4 ; X 2 = 19 cc — 10. Xj == 3 x- - 8 x - 1 ; Z 3 = 2025. The last step of the division may he omitted, for we only- wish the sign of X 3 , and that may he seen by inspection when — 104 x — 19 is obtained. We first substitute — go for x in each function, and obtain three variations of sign ; similarly + go gives no variation ; hence the three roots are all real. Substituting 0, we have two variations ; comparing this with the former results, we see that one root is negative and the other two are positive. The same result could have been obtained by Descartes' Rule, as all the roots are real. We now substitute various numbers to determine the limits of the roots. The table presents the results in a connected form : X *1 ^2 *3 When X = — 00, — + — + 3 variations. Li x = — 2, — + — + 3 variations. a x = — 1, + — + it x = 0, + — — + 2 variations. n X = 1, — + + a x = 2, — — + + 1 variation. a % == O} — + + + 1 variation. u x = 4, + + + a OC — Oj + + + + no variation. a X = GO, + + + + no variation. Then by Sturm's Theorem we know that there is one root between —2 and 0, one between and 2, and one bet ween 3 and 5. In fact, as X == when x = — 1, 1, and 4, these are the three roots of the equation. 2. Determine the number and situation of the real roots of X x 4 - 3 x s + 3 x- — 3 x + - = Note. In substituting the various numbers to determine the situation of the roots, it is best to work from o in cither direction, stopping when the number of variations is the same as has been previously found for + oo or — oo , as the case may be. SOLUTION OF HIGHER NUMERICAL EQUATIONS. 399 Here we find X, = 4x 3 - 9x 2 + 6x - 3 ; X a = — <d2x + 129 ; X = 3 x 2 + 18 x - 31 ; X i = - 1163. Substituting + co for x, we obtain one variation ; similarly, gives three variations, and — oo gives three variations. Hence there are only two real roots, both of which are posi- tive. We then substitute values of x from upwards, giving the following results : X ^ x 2 X s X 4 When x = 0, + — — + — 3 variations. ■ " x = l, + — — + - 3 variations. " x = 2, + + + — — 1 variation. " x = co, + + + — — 1 variation. Hence there are two roots between 1 and 2 ; and as the equation has four roots, there must be two imaginary roots. EXAMPLES. Determine the number and situation of the real roots of the following equations : 3. x*-x 2 -2x + l = Q. 6. x*-2x 9 -5x 2 + 10a;-3 = 0. 4. g* — 7x+7 = 0. 7. 2x- 4 -3* 3 + 17x- 2 -3x- + 15 = 0. 5. X s — 2 a -5 = 0. 8. x*- 4 X s -3 x + 27 = 0. XLIIL — SOLUTION OF HIGHER NUMERICAL EQUATIONS. 528. The real roots of the higher numerical equations in general can only be obtained by tentative methods, or by methods which involve approximation. Cubic and biquadratic equations may be considered as included in the class of higher 400 ALGEBEA. equations ; for their general solutions are complicated, and only of limited application. No general solution of an equa- tion of a degree higher than the fourth can he ohtained. COMMENSURABLE ROOTS. 529. A commensurable root is one which can be exactly expressed as an integer or fraction without using irrational quantities. An incommensurable root is one which can only he ex- pressed approximately by means of a decimal fraction. 530. Any equation containing fractional coefficients may be transformed into another whose coefficients are entire, that of the first term being unity (Art. 500), and such an equation cannot have a root equal to a rational fraction (Art. 495) ; hence, to find all commensurable roots, we have only to find all integral roots. 531. As every rational root of an equation in its general form is a divisor of the last term (Art. 493), to find the com- mensurable roots we have only to ascertain by trial what in- tegral divisors of the absolute term are roots of the equation. The trial may be made by substituting each divisor, both with the positive and the negative sign, in the equation ; or by dividing the first member of the equation by the unknown quantity minus the supposed root (Art. 486). In substi- tuting very small numbers, such as ± 1, the former method may be most convenient ; but when an actual root lias once been used, the latter method will give at once the depressed equation, which may be used in obtaining the other roots. 532. When the number of divisors of the last term is large, this process of successive trials becomes tedious, and a better method, known as the Method of Divisors, may be adopted. If a is a root of the equation x 4 +px i +qx 2 + tx + u = 0, SOLUTION OF HIGHER NUMERICAL EQUATIONS. 401 then a 4 + p a s + q a- + t a + u= 0. Transposing and dividing by a, -= — t — qa—pa 2 — a 3 , (1) u whence we see that - must be an integer. a Equation (1) may be written — h t = — q a —p a- — a*, cv U a a u Denoting - + t by t', and dividing by a, — — q —p a — a 2 , t' whence - must be an integer. a Proceeding in this way, we see that if a is a root of the equation, — \- t or t', — \- q or q', and \- p or p 1 must be in- Ct Ct Lt p' tegers, and 1-1 must equal zero. Hence the following RULE. Divide the absolute term of the equation by one of its inte- gral divisors, ami to the quotient add the coefficient of x. Divide this sum by the same divisor, and, if the quotient is an integer, add, to it the coefficient ofx 2 . Proceed in the same manner with each coefficient in regular order, and, if the divisor is a root of the equation, each quotient will be entire, and the last quotient added to the coefficient of the highest poicer ofx will equal 0. Equal roots, if any, should be removed before applying the rule ; and the labor may often be diminished by obtaining the superior limit to the positive and inferior limit to the nega- 402 ALGEBRA. tive roots of the equation, for no number need be tried which does not fall between these limits. 1. Find the roots of the equation x s - 6 x 2 + 27 x - 38 = 0. By Descartes' Rule, we see that the equation has no nega- tive root; and the only positive divisors of 38 are 1, 2, 19, and 38. By substitution we see that 1 is not a root of the equation. Dividing the first member by x — 2, we obtain x 2 — 4 x + 19 as a quotient. Hence 2 is a root, and the depressed equation is x 2 — 4 x + 19 = 0, from which we obtain x = 4±y/16-76 ~~2~ as the remaining roots. Hence, 2±v/-15 x = 2, or 2 ± y/ — 15, Ans. 2. Find the roots of the equation 8 x* - 4 x s - 14 x 2 + x + 3 = 0. We may write the equation A 9C i QC 0C O - *-2--^r + § + 8= - Proceeding as in Art. 500, we see that the multiplier 2 will remove the fractional coefficients. We then have the equation x 4 — x s — 7 x 2 + x + 6 = 0, (1) whose roots are twice those of the given equation (Art. 499). The divisors of G are ±1, ±2, ±3, and ± 6. By putting x equal to + 1 and — 1 in (1), it is readily seen that both are roots of the equation, and the other roots can be found from the depressed equation. But all of the rational roots may be obtained by the rule. 6, 3, 2, 1, -1, -2, -3,-6 1, 2, 3, 6, -6, -3, -2, -1 2, 3, 4, ("7 -5, -2, -1, 1, 2, 7, 5, 1, -6, -$ o, 9 -6, -7 -2, o, 2, 3 -3, -1, 1, 2 -1, -1, -1, -1 o, o, o, SOLUTION OF HIGHER NUMERICAL EQUATIONS. 403 It is customary to abridge the work as follows : Divisors, 1st Quotients, Adding 1, 2d Quotients, Adding — 7, 3d Quotients, Adding — 1, 4th Quotients, Adding 1, As 6, 2, — 3, and — 6 give fractional quotients at different stages of the operation, they cannot be roots of the given equation, and are rejected. 3, 1, — 1, and — 2 give entire quotients, and in each case the last quotient added to the coefficient of x 4 gives zero ; hence they are the four roots of 3 1 1 equation (1), and ^, ^, — ^, and — 1 are the four roots of the given equation. EXAMPLES. Find all the commensurable roots of the following equa- tions, and the remaining roots when possible by methods already given : 3. x s +Qx 2 +llx + Q = 0. 8. x s -7x 2 + 36 = 0. 4. x s + 3x 2 -4;x-12 = 0. 9. x 3 -6x 2 + 10a; — 8 = 0. 5. a- 4 -4.r 3 -8x + 32 = 0. 10. x 3 - 6 x 2 + 11 x- 6 = 0. 6. 4a; 8 -16sc 2 -9a; + 36 = 0. 11, 2x 3 -3x 2 +16x-2i = 0. 7. x 3 -3x 2 + x + 2 = 0. 12. x 5 - 2 x 3 -16 = 0. 13. x i -9x 3 + 23x 2 -20x + 15 = 0. 14. x 4 + x 3 - 29 x 2 - 9 x + 180 = 0. 404 ALGEBRA. RECURRING OR RECIPROCAL EQUATIONS. 533. A Recurring Equation is one in which the coef- ficients of any two terms equally distant from the extremes of the first member are equal. The equal coefficients may have the same sign, or opposite signs; hut a part cannot have the same sign, and a part opposite signs, in the same equation. Also, if the degree he even, and the equal coefficients have opposite signs, the middle term must be wanting. Thus, a 4 - 5 x z + 6 x 2 - 5 x + 1 = 0, 5 x 5 - 51 x* + 160 x 3 - 160 x 2 + 51 x - 5 = 0, x 6 — x 5 + x* — x 2 + x — 1 = 0, are recurring equations. 534. If any quantity is a root of a recurring equation, the reciprocal of that quantity is also a root of the same equation. Let x n +px n ~ 1 + qx n ~ 2 + ...± (... + </x 2 +i*z + l)=0 (1) be the equation. Substitute - for x ; then V y" y'" 1 y n ~- v v v > Multiplying each term by y n , (i+py+.qf.+,..)±(...+ qy n -*+py n - 1 +y n )=Q ( 2 ) Now, (1) and (2) take precisely the same form on changing the ± sign to the first parenthesis in equation (2), and hence they must have the same roots. Now, if a is a root of (1), as 1 1 v — - must be a root of (2) ; but, as (1) and (2) have the x a same roots, - must also be a root of (1). In like manner, if a b is a root of (1), y is also a root of (1). SOLUTION OF HIGHER NUMERICAL EQUATIONS. 405 On account of the property just demonstrated, recurring equations are also called reciprocal equations ; the former term relating to their coefficients, and the latter to their roots. 535. One root of a recurring equation of an odd degree is — 1 when the equal coefficients have the same sign, and +1 when they have opposite signs. A recurring equation of an odd degree, as x *n + l +px 2m + qx 2m-l + _ ± ( + q tf + p x + ^ _ Q (3) has an even number of terms, and may he written in one of the following forms, (x 2M + 1 + 1) +p (x 2m + x) + q (x 2 "'- 1 + x") + =0, ( X 2m + 1 — 1) +p (x 2m — x) + q (X lm - 1 -X-) + ...... = 0. If — 1 he substituted for x in the first form, or + 1 in the second, the first member will become ; hence, — 1 is a root of the first and + 1 a root of the second. If equation (3) be divided by x ± 1, both forms will reduce to the following form, X 2m +px 2 " 1 - 1 + qx 2m ~ 2 + + qx 2 +px + l = 0, (4) a recurring equation of an even degree in which the equal coefficients have the same sign. Hence, a recurring equation of an odd degree may always be depressed to one of an even degree. 536. Two roots of a recurring equation of an even degree are + 1 and — 1 when the equal coefficients have opposite signs. Let x *>» + pX 2m - 1 + qX 2m - 2 + -( + qx 2 +px+ 1)=0 be such an equation. As the middle term must be wanting (Art. 533), the equation may be written in the form ^™-l)+ F (^»" ! -l) + ?x 2 (f'»- 4 -l)-h = (5) which is divisible by both x — 1 and x + 1, or by x' 2 — 1 (Art. 120). Hence, both + 1 and — 1 are roots of the equation. 406 ALGEBRA. If equation (5) be divided by x 1 — 1, it will be depressed two degrees, and become a recurring equation of an even degree, in which the equal coefficients have the same sign (Art. 120). Hence, every recurring equation may be de- pressed to the form of equation (4), Art. 535. 537. Every recurring equation of an even degree, whose equal coefficients have the same sign, may be reduced to an equation of half that degree. Let a? m +qi x-"'- 1 + q ar m ~ 2 + + q x~ +p x + 1 = be such an equation. Dividing it by x m , we may write it (^ + ^) + ^(^"- , + ^)+?(^- 2 + ^r- 2 ) + ---=0 (6) the middle term if present becoming a known quantity. Put x + - = y Then, a 2 + -=- = y 2 - 2 x- x3+ x^ =y3 ~ s ( x+ x) =i/3 ~ s y x m + — = y m — m y m ~ 2 + Substituting these values in (6), we have an equation of the form y" l +2hy m - l + qiy m -* + = 0. After this equation is solved, we can immediately find x from the equation x + - = ?/. x 538. It thus appears that any recurring equation of the (2 m + l)th degree, one of the (2 m + 2)th degree whose equal SOLUTION OF HIGHER NUMERICAL EQUATIONS. 407 coefficients have opposite signs, and one of the 2 with degree whose equal coefficients have the same sign, may each he reduced to an equation of the with degree. EXAMPLES. 1. Given x 4 — 5 x s + 6 x 2 — 5 x + 1 = 0, to find x. Dividing by x 2 , far -\ — j j — 5 f x -\ — j + 6 = 0. S instituting y for x -\ — , and y 2 — 2 for x 2 -\ — - 9 we have x x l w 2 -2-5w+6 = 0. Whence, y = 4 or 1. If y = 4, x + - = 4, or x 2 — 4 x = — 1 ; Whence, x = 2 ± ^3. If y = 1, x + - = 1, or a; 2 — x = — 1 ; Whence, x = T Note. That 2 — y/3 and are reciprocals of 2 + y/3 and - — ^ 1 2 may easily be shown by reducing — — — and — - — ;=. to equivalent frac- 2 + ^3 1 + V— 3 tions with rational denominators (Art. 279). Solve the following equations : 2. a- 5 - 11 x l + 17 x 3 + 17 a- 2 - 11 x + 1 = 0. 3. a; 5 + 2 x* - 3 a; 3 - 3 x 2 + 2 x + 1 = 0. 4. x G — x 5 + x 4 — x 2 + x — 1 = 0. 5. a; 3 + px 2 +px + 1 = 0. 6. 6 x 4 + 5 a.- 3 - 38 x 2 + 5 x + 6 = 0. 7. 5 x 5 - 51 a,- 4 + 160 a; 3 - 160 a- 2 + 51 x - 5 = 0. 408 ALGEBRA. 8. x 4 + 5 x' + 5 x + 1 — 0. 9. x 5 = - 1, or x> + 1 = 0. (See Art. 332.) 10. x 5 -32 = 0. (Let cc = 2 ?/.) CARDAN'S METHOD FOR THE SOLUTION OF CUBIC EQUATIONS. 539. In order to solve a cubic equation by Cardan's method, it must first be transformed, if necessary, into another cubic equation 1 in which the square of the unknown quantity shall be wanting. By Art. 505, this may be done by substituting for x, y minus the coefficient of x 2 divided by 3. 540. If the first power of the unknown quantity be want- ing in the given equation, we may obtain the result by a simpler method, as follows : Let x 3 + a x 2 + c = be such an equation. Substituting - for x, we have V 1 a — + — + c = 0, or c y % + a y + 1 = 0. 541. To solve a cubic equation in the form x 3 + p x + q = 0. P Put x = z — =-, and the equation becomes Sz V 2 V 3 7) 2 or, z 3 - £—. + q = 0; or, 27 z* + 27 q z 3 -p 3 = 0. 2( z 3 This is an equation in the quadratic form, and may be solved by the method of Art. 313 ; and after z is known, x P may be found directly from the equation x = « — —-. o z SOLUTION OF HIGHER NUMERICAL EQUATIONS. 409 We have then for solving cubic equations the following RULE. If necessary, transform the equation into another cubic equation in which the square of the unknown quantity shall be wanting (Arts. 539 and 540). If y be the unknown quantity in the resulting equation, substitute for it z minus the coefficient of y divided by 3z. EXAMPLES. 1. Solve the equation x 3 — 9 x + 28 = 0. 3 Substituting z -\ — for x, 27 97 27 z 3 + 9z + — +^--9z - — + 28 = 0, z z 6 z 27 or, « 3 + tf + 28 = 0; or, * c + 28s 3 = -27. Solving by quadratics, z 3 = — 1 or — 27. Whence, z = — 1 or — 3. Uz = -1, x = z + - = -l-3 = -±. z If z = -3, x = - 3 - 1 = - 4. Hence, one root of the equation is — 4. Dividing the first member of the given equation by x + 4, we obtain as the de- pressed equation, x 2 — 4 x + 7 = 0. Whence, x — 2 ± y/— 3, the remaining roots. 2. Solve the equation x s — 24 x 2 — 24 x — 25 = 0. Putting x = y + 8 (Art. 539), we obtain 2/ 3 + 24^+192 Z / + 512-24y 2 -384?/-1536-24y-192-25=0, or, y 8 - 216 y — 1241 = 0. 410 ALGEBRA. 72 Putting y = z H — -, we have . _,_ 15552 373248 ._ 15552 , OM _ z z + 216s h 1 5 216 « 1241 = 0, z z* z or, z 3 + 3 ' 3 ^ 48 - 1241 = ; or, s c - 1241 z 3 + 373248 = 0. Whence, s s = 729 or 512, and 2 = 9 or 8. 72 72 Therefore, y = 9 + - r or 8 + -^ = 17, and sc = y + 8 = 25. 9 o Hence, one root of the equation is 25. Dividing the first member of the given equation by x — 25, we have as the depressed equation x 2 + x + 1 = 0: - 1 ± \J- 3 , Whence, x = - , the remaining roots. ii Solve the following equations : 3. x 3 - 6x + 9 = 0. 6. x 3 + 9 x 2 - 21 x + 11 = 0. 4. x 3 - 6 x 2 + 57 x -196=0. 7. x 3 -2 x 2 + 2x- 1 = 0. 5. :c 3 -4a; 2 -3x + 18 = 0. 8. x 3 -±x 2 + 4a;- 3 = 0. 9. a- 3 -3x 2 + 4 = 0. 10. Obtain one root of the equation x 3 + 6 x — 2 = 0. 542. In the cubic equation x 3 + px+q = 0, when p is — p 3 o 2 negative, and f, > =y , Cardan's method involves imaginary expressions ; but it may be shown in that case that the three roots of the equation are then real and unequal. Thus, in solving the equation x 3 — 6 x + 4 = 0. 2 Substituting z + - for x, we have z z 3 + 6 2 H \- -j — Gz f- 4 = 0, SOLUTION OF HIGHER NUMERICAL EQUATIONS. 411 r, g« + J* + 4 = ; or, z r > + 4 z* + 8 = 0. Whence, z z = -2 ± y/- 4, or -2 ± 2 y/- 1, or, ft = y/_ 2 + 2y/- 1 or ^-2-2^-1. It may be proved by trial that 1 + y/— 1 is the cube root of _ 2 + 2 yC3, and 1 - y/^1 of - 2 - 2 yCl. Hence, = l + y/-l or l-y/-l. If z = 1 + y/- 1, X=Z+-=1+ y/— 1 + 2 _2y/-l + 2 i + ^/=3 _ ' 1 + y'-i _ o Hence, one root of the equation is 2. Dividing the first member of the given equation by x — 2, we have as the de- pressed equation x 2 + 2 x - 2 = 0. Whence, £C = — 1 ± y^, the remaining roots. 543 We have no general rule for the extraction of the cube root of a binomial surd ; so that in examples like that in the preceding article, unless the value of z can be obtained by inspection, it is impossible to find the real values of x by Cardan's method. . In this case, the real values of x can always be found by a method involving Trigonometry. BIQUADRATIC EQUATIONS. 544. General solutions of biquadratic equations have been obtained by Descartes, Simpson, Euler, and others. Some of them require the second term of the equation to be removed, while others do not. All of them depend upon the solution of a cubic equation by Cardan's method, and will of course fail when that fails (Art. 542). They are practically of little value, especially as numerical equations of all degrees can be readily solved by methods of approximation. 412 ALGEBRA. INCOMMENSURABLE ROOTS. 545. If a higher numerical equation is found to contain no commensurable roots, or if, after removing the commen- surable roots, the depressed equation is still of a higher decree, the irrational or incommensurable roots must next be sought. The integral parts of these roots may be found by Sturm's Theorem or by Art. 517, and the decimal parts by any one of the three following methods of approximation. HORNER'S METHOD. 546. Suppose a root of the equation x" +j>x"~ 1 + q x n ~ 2 + + tx' 2 + ux + v = (1) is found to lie between a and a + 1. Transform the equation into another whose roots shall be less by a (Art. 502), and we shall have an equation in the form y n +p'V n ~ 1 + Q'y n ~ 2 + + t'y°- + u'y + v' = 0, (2) one of whose roots is less than 1. If that root is found to lie between the decimal fractions a' tenths and a' + 1 tenths, transform equation (2) into another whose roots shall be less by a' tenths, and we shall have an equation in the form s»+_p"2"- 1 + q"z n ~ 2 + ...... + t"z 2 + u"z + v" = (3) one of whose roots is less than .1. If that root is found to lie between the decimal fractions a" hundredths and a" + 1 hundredths, transform equation (3) into another whose roots shall be less by a" hundredths ; and so on. Thus we obtain x = a + a' + a" + to any desired degree of accuracy. As y and z in equations (2) and (3) are fractional, their higher powers are comparatively small ; hence approximate values of y and z may be found by considering the last two terms only, from which we have v' .. v" V = -77 and * ^ ~ TJ, ' SOLUTION OF HIGHER NUMERICAL EQUATIONS. 413 Thus approximate values of a', a", may be found in this way, and with greater accuracy the smaller they become. Hence a positive incommensurable root of the equation may be found by the following RULE. Find by Sturm's Theorem the initial part of 'the root, and transform the given, equation into one whose roots are less by th is initial part. Divide the absolute term of the transformed equation by the coefficient of the first power of the unknown quantity for the next figure of the root. Transform this last equation into another whose roots are less by the figure of the root last found, divide as before for the next figure of the root ; and so on. 547. A negative root may be found by changing the signs of the alternate terms of the equation beginning with the second, and finding the corresponding positive root of the transformed equation (Art. 498). This hy a change of sign becomes the-required negative root. 548. In obtaining the approximate value of any one of the quantities a 1 , a", by tbe rule, we are liable to get too great a result ; a similar case occurs in extracting the square or cube root of a number. We may discover such an error by observing the signs of the last two terms of the next transformed equation ; for, as the figures of the root as ob- tained in succession are to be added, it follows that a', a", must be positive quantities, so that the last two terms of the transformed equation must be of opposite sign. We then diminish the approximate value until a result is found which satisfies this condition. 549. If in any transformed equation the coefficient of the first power of the unknown quantity should be zero, the next figure of the root may be obtained by dividing the absolute term by the coefficient of the square of the unknown quantity, and taking the square root of the result. 414 ALGEBRA. For, if in equation (2), Art. 546, u' = 0, we have, approxi- mately, t' y 2 + v' = 0, whence y = \/ We proceed in a similar manner if any numher of the coefficients immediately preceding the absolute term reduce to zero. 550. 1. Solve the equation x 3 — 3 x 2 — 2 x + 5 — 0. By Sturm's Theorem, the equation has three real roots ; one between 3 and 4, another between 1 and 2, the third between — 1 and — 2. To find the first root, we transform the equation into another whose roots are less by 3, which by Art. 503 is effected as follows : Dividing x 3 — 3 x 2 — 2 x + 5 by x — 3, we have x 2 — 2 as a quotient and — 1 as a remainder. Dividing x 2 — 2 by x — 3, we have x + 3 as a quotient and 7 as a remainder. Dividing x + 3 by x — 3, we have 1 as a quotient and 6 as a remain- der. Hence the transformed equation is x 3 + 6 x 2 + 7 x - 1 = 0, whose roots are less by 3 than those of the given equation. Note. The operations of division in Horner's Method are usually per- formed by a method known as Synthetic Division. For example, let it be required to divide ofi - 19 x + 30 by x - 2. * 3 ±0.r 2 -19a; + 30 T 3 _ O r 2 x -2 a; 2 + 2x-15 2 a; 2 2.-C 2 - ix -15a; -15.Z + 30 The first term of each partial product may be omitted, as it is merely a repetition of the term immediately above. Also the remaining term of each partial product may be added to the corresponding term of the divi- dend, provided we change the sign of the second term of the divisor before SOLUTION OF HIGHER NUMERICAL EQUATIONS. 415 multiplying. Also the powers of x may be omitted, as we need only consider the coefficients in order to obtain the remainder. The work now stands l±0-19 + 30|l+2 + 2 | 1 + 2-15 + 2 + 4 -15 -30 As the first term of the divisor is 1, it is usually omitted, and the first terms of the dividends constitute the quotient. Raising the oblique columns we have the following concise form : Dividend, l±0-19 + 30|+2 Partial Products, +2+ 4-30 Quotient, 1 + 2-15,+ Remainder. Here we use only the second term of the divisor with its sign changed ; each term of the quotient is the sum of the terms in the vertical column under which it stands, and each term of the second line is obtained by multiplying the preceding term of the quotient by the divisor as written. By the method of Synthetic Division, the work of trans- forming the given equation into one whose roots are less by 3 stands as follows : 1 - 3 — 2 + 5| +3 + 3 - 6 — 2 — 1, 1st Remainder + 3 + 9 + 3 + 7, 2d Remainder. + 3 -f 6, 3d Remainder. Thus the transformed equation is, as before, x 3 + 6 x 2 + 7 x - 1 = 0. (1) Dividing 1 by 7 we obtain .1 as the next figure of the root, and we proceed to transform equation (1) into another whose roots shall be less by .1. 416 ALGEBRA. 6 7 -1 .1 .61 .761 6.1 7.61 -.239 .1 .62 6.2 8.23 .1 6.3 Thus the transformed equation is x s + 6.3 x 2 + 8.23 x - .239 = 0, whence by dividing .239 by 8.23 we obtain .02 for the next root figure ; and so on. Thus the first root is, approximately, 3.12. Similarly, the second root may be shown to be 1.201 ap- proximately. By Art. 547, the third root is the positive root of the equation x 3 + 3 x~ — 2 x — 5 — with its sign changed. The successive transformations are usually written in connection as in the following form, where the coefficients of the different transformed equations are indicated by (1), (2), (3), The work may also be contracted by dropping such decimal figures from the right of each column as are not needed for the required degree of accuracy. 1 3 _ 9 4 -5 | 1.33 1 2 4 2 (1)173 1 5 2.667 5 (1)7 (2)- .333 1 1.89 (1)6 8.89 .3 1.98 6.3 (2) 10.87 .3 6.6 .3 (2) 6.9 Hence, the third root is — 1.33 approximately. SOLUTION OF HIGHER NUMERICAL EQUATIONS. 417 EXAMPLES. Find the real roots of the following equations : 2. a r 3 _2x-5 = 0. 5. x 3 - 17 x 2 + 54 x -350 = 0. 3. a* + x 3 - 500 = 0. q a; 4 -4 a 3 -3 a: + 27 = 0. 4. x 3 - 7 a; +7 = 0. 7. x 4 - 12 x 1 + 12 x -3 = 0. APPROXIMATION BY DOUBLE POSITION. 551. Find two numbers, a and b, the one greater and the other less than a root of the equation (Arts. 517 or 521), and suppose a to he nearer the root than b. Substitute them separately for x in the given equation, and let A and B repre- sent the values of the first member thus obtained. If a and b were the true roots, A and B would each be ; hence the latter may be considered as the errors which result from sub- stituting a and b for x. Although not strictly correct, yet, for the purpose of approximation, we may assume that A : B = x — a :x — b Whence (Art. 348), A — B : A = b — a : x — a or (Art. 345), A — B : b — a = A:x — a (1) A(b-a) and, x — a = — — A — B A(b-a) or, x = a + A _ B . From (1), we see that, approximately, As the difference of the errors is to the difference of the two assumed numbers, so is either error to the correction of its assumed number. Adding this correction when its assumed number is too small, or subtracting when too large, we obtain a nearer approximation to the true root. This result and another 418 ALGEBRA. assumed number may now be used as new values of a and b, for obtaining a still nearer approximation ; and so on. It is best to employ two assumed quantities tbat sball differ from each other only by unity in the last figure on the right. It is also best to use the smaller error. This method of approximation has the advantage of being applicable to equations in any form. It may, therefore, be applied to radical and exponential equations, and others not reduced to the general form (Art. 480). EXAMPLES. 1. Find a root of the equation x z + x- + x — 100 = 0. When 4 and 5 are substituted for x in the equation, the results are — 16 and + 55, respectively; hence a = 4, b = 5, A — — 16, and B = 55. According to the formula,' the first approximation gives _ 16 (5 - 4) , 16 , _ * = 4+ -16-55 - 4 + 7l = 42+ - As the true root is greater than 4.2, we now assume 4.2 and 4.3 as a and b. Substituting these values for x in the given equation, we obtain — 4.072 and + 2.297 ; therefore 4.3 is nearer the true root than 4.2. „ , Q 2.297(4.3-4.2) .2297 Hence, x = 4.3- ^ + 4 _ = 4.3 - -^ -4.3 -.036 = 4.264. Substituting 4.264 and 4.265 for x, and stating the result in the form of a proportion, we have .0276 + .0365 : .001 = .0270 : correction of 4.264. Whence the correction = .00043+. Hence, x = 4.264 + .00043 = 4.26443+, Am. Find one root of each of the following equations : SOLUTION OF HIGHER NUMERICAL EQUATIONS. 419- 2. x 3 - 2 x- 50 = 0. 4. a; 3 + 8r+ Gx — 75.9 = 0. 3. * 3 +10x 2 +5a;-260 = 0. 5. x 3 +^-^- T = 0. lb 4 6. x 4 - 3 x 2 - 75 x - 10000 = 0. 7. a; 5 + 2 x 4 + 3 x 3 + 4 a; 2 + 5 x - 54321 = 0. NEWTON'S METHOD OF APPROXIMATION. 552. Find two numbers, one greater and the other less than a root of the equation (Arts. 517 or 521). Let a he one of those numbers, the nearest to the root, if it can be ascer- tained. Substitute a + y for x in the given equation ; then y is small, and by omitting y 2 , y 3 , , a value of y is obtained, which, added to a, gives b, a closer approximation to the value of x. Now substitute b + z for x in the given equation, and a second approximation may be obtained by the same process as before. By proceeding in this way, the value of the root may be obtained to any required degree of accuracy. The assumed value of x should be nearer to one root than to any other, in order to secure accuracy in the approxima- tion. EXAMPLES. 1. Find the real root of the equation x 3 — 2 x — 5 = 0. When 2 and 3 are substituted for x in the equation, the re- sults are — 1 and + 16 respectively ; hence a root lies between 2 and 3, and near to 2. Substitute 2 + y for x, and there results y 3 + 6y" + 10y-l = 0. Whence, approximately, y = .1. Now substitute 2.1 + z for x, and there results .061 + 11.23 2 + = 0. 420 ALGEBRA. Whence, approximately, z = ' i}0 = — .0054, and x = 2.1 - -0054 = 2.0946, nearly. Find one root of each of the following equations : 2. a; 8 — 3 x -(- 1 = 0. 3. x s - 15 x- + 03 x - 50 = 0. ANSWERS TO EXAMPLES. In the following collection of the answers to the examples and problems given in the preceding portion of the text-book, those answers are omitted which, if given, would destroy the utility of the problem. Art. 47; pages 10 and 11. 1. 93. 5. 408. 9. 5§. 13. 36. 17. llf. 2. 136. 6. 254. 10. 13$. 14. 48. 18. 9. 3. 127. 7. 24. 11. 4. 15. 3. 19. 10. 4. 156. 8. 310. 12. If. 16. 4. 20. 76. Art. 60; page 18. 6. Ua-9vip 2 . 7. x. 8. Sab-lcd. 10. 3mn 2 -2x 2 y. 11. 39 a 2 -2i ab + 5 h\ 12. - « + 3 c + 2. 13. x-y+3m + 3n. 14. 3a + 3b + 3c + 3d. 15. x. 16. n+ r. 17. 6mn — ab — 4 c + 3x + 3m 2 — 4 p. 18. 4 a - 2 b - 12 - 3 c - d + 4 x 2 - 18 m. 19. 6 a 3 . 20. 14 six. 21. 7 a b + 7 (a + b). 22. 16 \/ y - 4 (a- b). Art. 66; page 21. 6. — 3a5+4ctf— 5 a x. 7. 6 z + 12 ?/ — 8 « + 4. 8. _4a6c-14a--2y- 148. 9. 2 y/ a - 4 y- + 12 a .+ 1. 11. 14 x 2 -Sj/ 2 + 5ab-7. 12. 2 b -2 e. 13. 6 b + 1. 14. 4m^8»-r+3s. 15. Qd -2 b — 3 a -3c 16. 5m 2 +9w 3 -71a;. 17. 2 b. 18. a -6 -3 c. 422 ■ ALGEBRA. Art. 74; page 24. 4. a — b + c + cl — e. 5. 2 a + 2. 6. x — y. 7. a — 3b + c. 8. 5wr-6»-4(i. 9. 6?» — 3%. 10. 4x + 2y. 11. —35 — 7c. 12. a — c. 13. 9a+l. 14. 6 m + 2. Art. 86; pages 29 and 30. 3. 6a 3 -16a 2 ?/+6ay 2 + 4?/ 3 . 4. a 4 + 4a + 3. 5. a 2 -b 2 + 2bc-c\ 6. -0a 2 +16a&-8£ 2 . 7. 6 8 -a 8 . 8. o 4 -«. 9. 30 a 3 - 43 a 2 b + 39 a b 2 - 20 b 3 . 10. 6 a 4 + 13 x 3 - 70 a 3 + 71a-20. 11. -a 5 - 37a; 2 + 70 x -50. 12. -Gar 5 - 25 a 4 + 7a 3 + 81a 2 + 3a-28. 13. 2a 5 b 2 -3a 4 b 3 -7 a 3 b 4 + 4:a 2 b s . 14. 4x 2m+1 y 3 — 1G x m+6 y n + 1 + 12 x 5 y 2n ~\ 15. 12 a; 6 +7 a; 4 + 5 x 3 + 10 x — 4. 16. m 5 + re 5 . 17. a 5 — 5 a 4 6 + 10 a 3 b 2 -10 a 2 b 3 + 5 ab 4 - b\ Art 87; page 30. 2. 6 a 2 + 11 a J + 4 6 2 . 3. a 5 + x 5 . 4. a 8 - 2 a 4 a 4 + x 8 . 5. 2 a m + 1 - 2 a n + 1 - a m + n + a 2n . 6. 1 - a 8 . 7. a 3 + 3 a 2 a - 10 a x 2 - 24 x 3 . 8. a 5 -5 a 4 + 10 a 3 -10 a 2 + 5 a- 1. Art. 101; pages 37 and 38. 3. a x - 2. 4. 3 b 2 - 4 a 2 . 5. 4 « 2 - 3 b 2 . 6. 3« 4 + 3a 3 i + 3a 2 6 2 + 3«6 3 + 36 4 . 7. a 2 -ax + x 2 +-^—. 8. x 3 -x 2 y a + x + xy 2 -y 3 + —¥—. 9. 2 x 2 - 7 a - 8. 10. 5 x 2 - 4 a + 3. a; + y 11. x 2 - 2 x - 3. 12. a 4 + x 3 y + a- 2 ?/ 2 + x y 3 + y\ 13. 3a 3 -2a 2 + a-5. 14. 2 a- 3 - a- + 1. 15. a-b+c. 16. a 2 — 3 a— y. 17. x + //. 18. a n — b m + c r . 19. l + 2a + 2a 2 + 2a 3 + ... 20. a-ax + ax 2 -ax 3 + ... ANSWERS TO EXAMPLES. 423 21. a*-an + a i b 2 -ab* + b i . 22. 2 a 3 - 2 a 2 - 3 a - 2. 23. - x 2 - 2 x - 4. 24. a 3 - a: + 2. 25. 2 a 2 - a b + 2 & 2 . Art. 107; page 40. 23. 1 - a 2 + 2 a b - b 2 . 24. a 2 - b 2 - 2 b c - c 2 . 25. a 2 -2ab + b 2 -c 2 . 26. c 2 - a 2 + 2 «6 -£ 2 . 27. a 2 + 2«i + i' 2 -c 2 + 2^- c/ 2 . 28. a 2 - 2 a 5 + 5 2 -c 2 +2crf-£ 29. a 2 + 2 a & + & 2 - c 2 - 2 c d - d 2 . Art. 115; page 42. 3. ( a + x )(b + y). 5. (x + 2)(x- y). 7. (x 2 - y 2 ) (m - n). 4. (a-m)(c + tf). 6. (a-b)(a 2 + b' 2 ). 8. (a: + 1) (x 2 + 1). 9. (3 a; + 2) (2 a; 2 -3). 12. (ab-cd) {ac + b d). 10. (2 a; - 3 y) (4 c + d). , 13. (m 2 x-ny) (n 2 x - m y). 11. (2-7m 2 j(3»-4m). 14. (4ww-7a;y)(3a& + 5crf)- Art. 117; page 45. 9. (a + b + c + d) (a + b — c — d). 10. (a — c + b) (a — c — b). 11. (m + x—y) (m — x + y). 12. (x — m + y — n) (x — m — y+ n). 16. (x + y + 2)(x + y-2). 18. (3c + d+l)(3c + d-l). 17. (a + b-c)(a-b + c). 19. (3 + x 2 -2y) (3-x 2 + 2y). 20. (2 a — 5 + 3 d) (2 a - J - 3 d). 21. (2 m 2 + 2 b - 1) (2 m 2 - 2 b + 1). 22. (a — m + b -f- n) (a — vi — b — n). 23. (a + m + b — n) (a + m — b + n). 24. (x — c + y — d) (x — c — y + d). Art 118; page 49. 25. (x 2 - 24) (x 2 - 5). 27. (x y 3 + 12) (x y 3 - 10). 26. (c 3 + 11) (c 3 + 1). 28. (a b 2 - 16) (a b 2 + 9). 424 • ALGEBRA. 29. (x + 20 n) (x + 5 n). 32. (x + y - 5) (x + y — 2). 30. (m 2 + 11 n 1 ) O' 2 - 6 ir). 33. (x - 8 y 2 g) (aj + 6 y 2 s). 31. ( a _ 5 _ 4) ( a _ 5 + 1). 34. (m + n + 2) (m + ra - 1). Art. 121 ; page 53. 3. Sab (a -' r 2) 2 . 7. 3 a 2 (a - 5) (a - 2). 4. 5 x y 2 (3 x - 4 y 2 ) 2 . 8. 2 c m (c + 7) (c - 3). 5. 2ay(3a; + y)(3a; — y). 9. a; y (m — 6) (m + 2). 6. a; (x + 7) (a; + 1). 10. ±ab(2a + b) (±a 2 -2ab + b 2 ) 11. (n - 1) (>r + » + 1) O 6 + w 3 + 1). 12. (x 2 + y 2 )(x-t- 2 /)(^- 2 /). 13. (cc 4 + m 4 ) (a; 2 + m 2 ) (x + m) (x — m). 14. (m + w) (??i — w) (m 2 + m w + ?z 2 ) (??i 2 — mn + ri 2 ). 15. (a + c) (a 2 - a c + e 2 ) (a 6 - a 3 c 3 + c 6 ). 16. (2 a + 1) (2 « - 1) (4 a 2 + 2 a + 1) (4 a 2 - 2 a + 1). Art. 125 ; pages 55 and 56. Z. ax. 6. x + 7. 9. x (x — 1). 12. 2 a: + 5. 4. m + n. 7. 2 a; — 3. 10. a — 2 &. 13. m(j;- 1). 5. x 2 + 1. 8. 3 a - 4. 11. x + 6. 14. 4 a? - 1. Art. 126 ; page 61. 6. 2 x + 3. 10. 2 a; - 5. 14. x 2 + x + l. 7. 8a; — 7. 11. 5x + 3. 15. a — a;. 8. x - 1. 12. jb + 2. 16. x 2 - 2. 9. 3 a; + 4. 13. 2 a; - 1. 17. 2 (x + y). 18. 2«-3x. 19. 3 a; + 2. Art. 130; page 63. 2. 120a 4 6 2 c. 4. 36 a & b\ 6. 840 a 2 c 2 d 3 . 3. 30xV« 8 . 5. 480 m 3 rt 2 x 2 t/ 2 . 7. 252 a 8 y 3 s 8 . 8. 1080 a 2 b 2 c 3 d\ 9. 168 m n 2 x z y s . ANSWERS TO EXAMPLES. 425 Art. 131; pages 63 and 64. 2. ax(x-\-a)(x—a)(x-+ax+a 2 ). 7. ax (x — 3) (x— 7) (x + 8). 3. 12 abc (a + b)(a-b). 8. (2x+l)(2x-l)%4:x 2 +2x+l). 4. «(cc + l)(a;-l)(x 2 — rc + 1). 9. 3 ab (x — y) 2 (a — b). 5. 24(l + cc)(l-x)(l + a: 2 ). 10. 2az; 2 (3;c + 2) 2 (9a; 2 -6x+4). 6. (x+l)(a;-2)(cc+3)(*+4). 11. (x-l)(x-3) (x + 4)(x-5). 12. (x + y + z) (x + y — z) (x — y + z). Art. 132; page 65. 2. (3x-4)(4 : r-5)(2a; + 7). 4. (a 2 -2a-2)(a + 3)(2a-l). 3. (4a;+l)(2a;+7)(3a;-8). 5. (2x + 3)(x 2 -x + l){x 2 +x-2). 6. (a — b)(a 2 -ab + b 2 )(a 2 + 2ab + b 2 ). 7. a x (x + 1) (x 2 — x — 1) (x 2 + x + 1). 8. x (x - 5) (2 x 2 - x -2) (3 x 2 + x - 1). If the above expressions are expanded, the answers take the following forms : 2. 24 z 3 + 22 a 2 -177 a; + 140. 4. 2« 4 + a 3 -17« 2 -4« + 6. 3. 24 z 3 + 26 z 2 - 219 a; -56. 5. 2x 5 + 3x i -±x 3 + 5x-Q. 6. a 5 -a 3 b 2 + a 2 b 3 -b\ 7. a x G + ax 5 — ax 4 — 3 ax 3 — 3 ax 2 — ax. 8. 6ic 6 -31a; c -4x 4 + 44a; 3 + 7a; 2 -10a;. Art. 148 ; page 71. 14 - *£■■ 18 - o + 2 c a (2 + 3 n) 1D * b{2-3n) • 19 ' 16. ±*--J*y+v\ 20 . m + 9 3 y — 5 10. cd 3 xy' 11. x a 2?' 12. sc — 5 # + 7' 13. m — 2 2 x 2 y 5 — X 2 + x X c (J-x) -d c III + d' . — nr m 2 — n 426 ALGEBEA. Art. 149; page 72. 2. 3 * 3. 4 a; + 1 5a + 7 a-2 ' 10. 4. m — 1 6 m — 5 5. x + 2 x — 3 " x' 2 3a; + l x' z — x + 3 Art. 150; page 73. 6. 3 a; — 2 8 2a; + 5 2 x -7 a; + 3 * 7. 2x-3 _ C « — 1 "• 7= » 5a — 7 2a;-5* , 1 2x 2 -a;-2 U< 2 a 2 + 3a; + l* 3 a - a* 4. a; 2 — X !/ + */• 6. a; 2 3" a; ~~3 7 + 3" 2 a;" 7. a 2b' 3 2b 2 + a ' K 2x 3 o 5 5 5a; 23 8. 2a; + 6 + 9. x 2 + a; + l. 10. 2 + 77 -^- — ^-, 11. a;-24 12. jc x — 3 2a;-4 2x z — x+1 ,x z +x — 1 a; — 2 2a; 2 -3a; + 3' Art. 151; page 74. _ (x — 1) 2 _ an + b-—cd 56.r — 4?i 2 — 5a 2. =— . o. . 4. - . x — 6 n o (a; + l) 2 . 2ab „ a 2 +2b 2 Q a 3 + b 3 0. . b. j. 7. — s . o. — . x a+b 2 a a — b Q 6a; 2 -7a;-l 2b 2 ,, a; 3 -2a; 2 -3a; "• ^ : n • I". — ■ -. 11. „ . 2a; — 1 a+& x — 2 Art. 152; pages 76 and 77. 27 a b 16 a c 30 6 n 3 a; 2 ?/ 2 a-?/,? 7 y z 2 3 ' 772~' "T^ 7 ~72~" 30 ' 30 ' 30 ' 18 ?/a 2 16 x 2 z 2 15 a; 2 y 2 ' 12 xyz* 12xij z' 12 xyz' 40c 2 -10c 18i 2 -12 6 25 « a _2^_ 3_o^ 4 a a; 2 30 a&c ' 30 abc ' 30 a be' ' a*x 8 ' a 8 a; 8 ' a 8 a; 3 ' 8. ANSWERS TO EXAMPLES. 427 100 a y z 3 Aob x 3 z 8Acxy s — 12mx y 2 120 x 1 ?fz 2 ' 120 x 2 y 2 z 2> ' 120 x 2 y 2 z 2 * (a+b) {a 2 + U 1 ) (a - b) (a 2 + b 2 ) a 2 - b 2 10. a 4 -6 4 at-b* ' at-W x 2 — 9 x 2 -l x 2 -4 (x-l)(x-2)(x-3)' (x-l)(x-2)(x-3)' (x-l)(x-2)(x-3)' 2 a (a + 2) 3 b (a — 2) 4 c (a + 3) 12. 0-2)(a+2)0+3)' 0-2)0+2)0+3)' (a-2)(a+2)(a+3)' x 3 +2x 2 -\-2x + l x 2 + x + l x + 1 (x + 1) {x 3 - 1) '0 + 1) (X s - 1)' {x + 1) (a 3 - 1)' 13 6 a 2 b 2 3 b (m 2 - n 2 ) 2 a (a 2 - b 2 ) 6ab(a — b)(m + ?iy 6ab(a—b)(m + n)' 6ab(a—b)(m + ii)' 3 Q+l) 2Q-1) 2-a 1-x x 2 -x-2 3 ' a»-l' a 2 -l'a 2 -l' "*' l-x» 1-jb 2 >1=?' 17 c2 -^ 2 Q-l)Q + &) Q-6) 2 ' O 2 - 6 2 ) (c - d)' O 2 -* 2 ) (e-d)' (a 2 -b 2 ) - d)' Art. 153; page 78. 2 ( a ~ b y 3 g 2 +9«+8 9 m 2 - 4 a* — b 2 ' ' x 2 +5x — 24' "6m 2 — 19m + 10" 4 2 +«& + 6 2 ) 1_^ a 3 — & 3 1 — x Art. 154; pages 80 to 82. 4 12 a; + 7 6a + 5& a + 3 3 m 2 n 2 - 4 36 ' ' 10 a 2 6 2 ' 24 • 6m 2 -» 3 * 8 5 &* + 4 «' q 5a + 6 in 4,ab-b-±a 3 1 " 120 a 6 " * 24 • iUl 12« 3 T~ ~* 1L 15" 12 — is 3a; ~2 1 45ctf+6acd— 3ahd— 2abc •42' "■ 18 x 2 ' 14l_ 60- 15 -~ ~48^X~ 17. ? 18 1 19 2 Q^ + & 2 ) 6 + x-x 2 ' ' x 2 +15x+56' a 2 -b 2 ' 428 ALGEBKA. ■Oi *" «l£±|. 22.. 4^. 23. (* + 2 y 1_^' — a-6* -a 4 -^' • (.x + l)(a; 3 -l) 13-18* _J_ a 2 -14a + l ^ (z + lXa^Xz-S)* 0, &-«' ' 6(^-1) 28. 57 _?-_. 29. 44- 30. 0. 31. 9 a:- a; 3 ' ' »*— 1" ' ' ' (»-2)(jb-3)(»-4) Art. 155; pages 83 to 85. 2. a 5 b s c m* n 3 d' 3. 12 a 4 b x 35 h 5 m ' 4 ^ 7. / • 8 - 4 a.' y a 3 . 9 11 o m 4 w 2 3 a; a; - -1 10* -2 * "■ 3 13. b(a-b) x(a + b) 14. a — b a 2 ' I*' 1 -'. a; ia x*-x-2Q 16 ' 17. a (x-2) a + 1 18. X x-2' 19. "+*• . or 20. aj2 + 5 f + 6 ST 21. -^o . 22. x- + xy. 23. 1. 24. 2. 25. -J£-j. ic + 2 % ~r y . 91 ra 2 o. 6rc 2 2 4. Art. 156 ; page 86. 5 mx 7.!. X 9.» + 1 ' a + 5 , 3 0-4) 8 3.t- 2?/ 10. aj. 8 b 3 m ri 1 x s x + y Art. 157; pages 88 and 89. a _ acn + bn „ 3m— n „ Ay—±x + 2a — — . o. — — . b. — x . «. ^j • bm + b?i cnx — cm ox oi x 2 y 3 +l 8. x — 1. 9. a; 2 -rK + l. 10. a + b. 11. xif — ly* , n a— J 10 «— 4 ... 1( . 4 R aft 12. -T7-. 13. ^. 14. aj. 15. k— -Q- lb - ■> , 7,2 ■ a- b x + 6 3 a; + 3 a" + ANSWERS TO EXAMPLES. 429 17 1 18 ww(m-w) 8 19 _ s-g m* + m 1 ri 1 + w 4 ' x + 2 a Art. 175; pages 94 and 95. 2. aenx—bcen=bdnx—bem. 3. 6 bx — 8a 2 = 3 — 2a6a. 4. bdex— adex+bcex— abd=0. 5. \2x-\-hx — 6x— 1320- 6. 9z-12a=10a:+24-46. 7. 28a:-4a;+560=14a;+7a;+728. 8. ±ax-6c-5a s x + 2a 3 bd=0. 9. 10x-32a;-312=21-52a;. 11. 3 a — 2 a — 2 a; = 45. 12. a 6 a; + b 1 — ex — d = a c. 13, 3-3z-2-2x = 0. 14. 6ar+3z-6ar+ 18-4a;-2=0. 15. 3 a ._3_2a--2-5x=0. 16. 6x + 6-15aj + 45-20a;-10=0. Art. 177; pages 97 to 102. 4. 3. 13. 1. 22. 72. 31. 5. 41. - 1&. 5. 7. 14. 2. 23. 60. 32. -5. 42. 4£. 6. -1. 15. 2. 24. 10. 33. 4. 43. l t V 7. 16. -4. 25. -2* 34. -5. 44. 0. 8. 1. 17. 2. 26. 56. 35. -2. 45. H. 9. 3 < 18. 1. > 27. ?. 33. f . 46.-^. 11. 3 — 2' 20. 3. 29. -2. 37. -1. 47. 1. 12. 0. 21. 5. ao.J. 40. -7. 50. f -* 2a+b 51. a 2 a 2 - + 4a . 52 3a+2 !. 2&. 53. If *I Abc + a* a?—b + 16c 55. 2 a 2 36 , 57. ^. 59. \ a + 2 i- 61 --^ 64. -3. 56. a. r 58. 12 a 3 . 60. aft. 63. 2. 65. 50. 66 B5- 67. 5. 68. 0. 430 ALGEBEA. Art. 182 ; pages 108 to 113. 10. Horse, $224; chaise, $112. 11. 37. 12. 10 and 7. 13. 18 and 2. 14. 58J and 4l£. 15. A, 40 ; B, 20. 16. A, 60 ; B, 15. 17. If. 18. ft. 19. 23£. 20. 84. 21. 36. 22. Oxen, 12 ; cows, 24. 23. Wife, $864; daughter, $288; son, $144. 24. Worked, 20 ; absent, 16. 25. Horse, $ 126 ; saddle, $ 12. 26. Infantry, 2450 ; cavalry, 196 ; artillery, 98. 27. 144 sq. yds. 28. Water, 1540 ; foot, 880 ; horse, 616. 29. $ 1728. 30. $ 2000 at 6 p.c. ; $ 1200 at 5 p.c. 31. 7. 32. 31. 33. $24. 34. $100. 35. 142857. 36. A, $ 466| ; B, $ 533J. 37. 2 dollars, 20 dimes, 4 cents. 38. $2.75. 39. Men, $25; women, $21. 40. 23 and 18. 41. 48 minutes. 42. 12121 men ; 110 on a side at first. 43. 5 r \ minutes after 7. 44. 43 T 7 T minutes after 2. 45. 27f\ minutes after 5. 46. 29 and 14. 47. 3377 ounces of gold ; 783 ounces of silver. 48. $ 2000. 49. 30 bushels at 9 shillings ; 10 at 13 shillings. 50. 10 a.m. 51. $ 1280. 52. 21 ft minutes, or 54 T <Y minutes after 7. 53. 27^ T minutes after 4. 54. 23*1 miles. 55. Greyhound, 72 ; fox, 108. 56. 1 minute, If §f seconds. Art. 192 ; pages 120 to 123. 3. x = 4, y = 3. 9. x=-2, y=10. 15. x = 12, y = 18. 4. s = 5, y = -2. 10. jk==12, y = S. 16. a; = 35, y = -10. 5. x = 7, y = 5. 11. x=~2, y=-10. 17. x = - 28, y = 21. 6. x = - 8, y = 2. 12. x = 10, y = 5. 18. x = A,y = .1. 7. x = 5, y = 7. 13. x = 7, y = 11. 19. x = l£, y = 3f . 8. x=-8, y=-12. 14. a>=ll, y=-9. 20. x = 3, y = -2. 21 dm — bn _an — c m n / + ri r ad — be ad — be ~ mri + m' n ' ANSWERS TO EXAMPLES. 431 m'r—mr' nn ac(bvi + dv) bd(cn-am) V = • "3. X = ^-= ; -, y = ^ : -. u m n' + w! n ad + b c ad + b c 11 25 5* 24. x=—, y=-^. 25. x = 60, y = 40. 26. x = — , y=». 2 a 2, a Ob 27. x = lft, y = 4 T V 28. x = -6, y=-5. 30. # = 4, y=2, be — ad bn — d vi ' 31. x = — 5, y=3. 32. x = -2, y — — l. 33. ■ x = b c — a d „. 3 2 oe 1 1 y = • 34.33 = ^77,2/ = —^. 35. x=-,y = — cm — an a~ b a tr n vi Art. 194; pages 126 and 127. 3. x = 23, y = 6, g = 24. 6. a; = — 5, y = — 5, z = — 5. 4. x. = — 2,y = 3, z = l. 7. « = 4, x = 5, y = 6, z = 7. 5. x = 8, y = - 3, g = - 4. 8. a; = 3, y = - 1, s = 0. 9. jc =^ (b + g — a), y = ^ (a + c — b), y = ^ (a + & — c). 10. x~^,y = 7 -£,z=^ 11. a = -24,y=-48,g=60. 12. u = — 7, x = 3, ?/ = — 5, z = 1. _ 5 2 +c 2 -ft 2 _a 2 +c 2 -& 2 _ a 2 + & 2 -c 2 13, *~ 2bc ~ ,V ~ 2ae ,Z ~~ 2ab ' 14. x = -,y = -^,z=--. 15. a; = li y = -X\, «='l. 16. x = ab c, y=ab + ac + bc, z = a + b + c. a + 1 f- 17. x = 7, y — — 3, g = — 5. 18. as = ,y—a — c,z-- c a Art. 195 ; pages 129 to 133. 4. A, 30 ; B, 20. 5. ^. 6. Cows, 49 ; oxen, 40. 7. A, $140; B, $70. 8. A, 98; B, 15. 9. 32 and 18. 10. Man, 24 ; wife, 18. 11. Worked, 6 ; absent, 4. 12. Horse, $96; chaise, $112. 13. A, $96; B, $48. 432 ALGEBRA. 14. 16 days. 15. 13J bushels at 60 cts. ; 26| at 90 cts. 16. Wheat, 9; rye, 15. 17. Income tax, $20; assessed tax, $30. 18. A, $500; B, $700. 19. 30 cents ; 15 oranges. 20. 1st, 8 cts. ; 2d, 7 cts. ; 3d, 4 cts. 21. Better horse, $40 ; poorer, $30; harness, $50. 22. 10, 22, and 26. 23. 246. 24. A, $2000; B, $3000; C, $4000; D, $5000. 25. A, 45; B, 55. 26. A, $20; B, $30; C, $40. 27. Whole sum, $120; eldest, $40; 2d, $30; 3d, $24; 4th, $26. 28. Length, 30 rods ; width, 20 rods ; area, 600 sq. rods. 29. Going, 4 hours ; returning, 6 hours. 30. A, 9f days ; B, 16 ; C, 48. 31. 1st rate, 6 p.c. ; 2d, 5 p.c. 32. 15 miles ; 5^ miles an hour. 33. 30 miles an hour. 34. A, 5 ; B, 6. 35. First, 22 ; second, 10. 36. A, 8 ; B, 6. Art. 197; pages 136 and 137. . a b c _ . , _ m a .. n a 4. — — . 5. li hours. 6. and . ab + ac + bc vi-\-n m,-\-n 7. 12 and 8. 8. -^-. 9. 12. 10. 10 °" 11. $2100. b — a rt+ 100 12. 100 >-i>), 13. m. 14. lst, g(c - 6) ;2d ; ft(a - r) . p r ~ a — b a — b ,.-.». ^,.,/n , „ b + d ^„ am+bji + cp 15. 1st kind, 5 ; 2d, 10. 16. — — . 17. - -r- — . a — c a + b+c * amt . -o ant ' n a P t " mt + nt' -\-pt' n ' mt + nt'+ptf n ' mt+nt'+pf' Art. 205; page 141. 5 3.-2 rods. 4. - ^ . 5. 105 and — 15. 6. In — 30 years. 7. A, .-$1500; B, -$500; that is, A was in debt $1500, and B $ 500. 8. Man, $ 3 ; son, — $ 0.50 ; that is, the man was at an expense of 50 cents a day for his son's subsistence. ANSWERS TO EXAMPLES. 433 Art. 225; page 152. 4. x > 5. 5. x > 15, x < 20. 6. 4. 7. x > 6|, y > 2|. 8. a; > c, x < d. 9. * > 9|, 7/ < 12£. 10. 19 or 20. 11. Any no,, integral or fractional, between 8 and 15. 12. 60. Art. 229; page 155. 2 7 2 1. a 3 -3a 2 b + 3ab 2 -b 3 . 2.~-2+— i . 3. 1 + 3 a 2 + 3 b 2 + 3 a 4 + 6 a 2 6 2 + 3 b* + a & + 3 a 4 b 2 + 3 a 2 6 4 + 6 6 . 4. a 2 + 2am — 2an + m 2 — 2mn + n 2 . 5. a *m _ 4 a 3m + », _|_ (5 a 2m + 2n _ 4 (( m + 3a + a 4 n# 6. a 5 + 5 a 4 5 + 10 a 3 b 2 + 10 a 2 b s + 5 a 6 4 + 6 5 . Art. 230 ; page 156. 3. 4a: 4 +12ar+25ar 2 +24z + 16. 4. 4a: 4 -12a; 3 + lla; 2 -3a: + |:. 6. a; 6 +4a: 5 +6a; 4 +8a; 3 +9a: 2 +4a;+4. 7. l-4a;+10a: 2 -12a; 3 +9a: 4 . 8. 1 + 2 a; + 3 ar + 4 x 3 + 3 a; 4 + 2 a: 5 + a: 6 . 9. x 6 -8x 5 + 12 x* + 10 a; 3 + 28 x 2 + 12 x + 9. 10. 4 x« + 4 a: 5 + 29 x" + 10 a; 3 + 47 a: 2 - 14 x + 1. 11. a; 6 + 10 x 5 + 23 x i - 6 a: 3 + 21 x 2 - 4 a; + 4. 12. 9 a; 6 - 12 x 5 - 2 x i + 28 x 3 - 15 a; 2 - 8 a: + 16. Art. 231; page 157. 2. a^+Qa'b + 12 a 2 b 2 +U 3 . 3. 8m 3 +60?7r?z+150mw 2 +125» 3 . 4. 27 a; 3 -108 a; 2 +144 a; -64. 5. 8 a; 9 - 36 a; 6 + 54 x 3 - 27. 6. 64 x 6 -48 x 5 y +12 a; 4 ?/ 2 — a; 3 ?/ 3 . 7. 27 a; 3 ?/ 3 + 135 a b 2 x 2 y 2 + 225 a 2 b A xy + 125 a 3 b 6 . Art. 232; page 158. 3. a; 6 - 3 x 5 + 5 a; 3 - 3 x - 1. 5. 8 - 24 a; + 36 x 2 - 32 a: 3 + 18 x i - 6 a; 5 + a; 6 . 6. 1 + 3 x + 6 a; 2 + 10 a; 3 + 12 a; 4 434 ALGEBRA. + 12 x 5 + 10 x« + 6 x 1 + 3 x 8 + x\ 7. 8 x» - 12 x s + 30 a; 7 - 61 x* + GG x 5 - 93 a 4 + 98 a; 3 - 63 a 2 + 54 a; - 27. Art. 239 ; pages 162 and 163. 2. 2 a; 2 - x - 1. 5. 3 - 2 a; + a; 2 . 8. 3 ar - 4 x - 5. 3. 2 a 2 - 4 a + 2. 6. 5 + 3 x + x\ 9. 2 x 2 - 5 a; + 8. 1 4. m + 1 m 7. 1 — 7» -2 a; 2 . 10. a -b c. U. x-2y + 3z. 13. a-+ S fr—, : 4 a; a; 2 a; 3 i 3 8 a 3 16 a 5 15. a + s 12 - 1 + 2-^ + 16 2 a 8 a 3 + x° 16 a 5 Art. 241; page 166. 2. 523. 7. 95' 12. 900.8. 17. 13.15295 3. 214. 8. 1.082. 13. .4125. 18. .88192. 4. 327. 9. 21.12. 14. 1.41421. 19. .43301. 5. 5.76. 10. .083. 15. 2.23607. 20. .57735. 6 .97. 11. .00328. 16. 5.56776. 21. .53452. 1. 3.3166. 2. 1.732051. Art. 242; page 168. 3. 7.81024968. 5. 27.94638. 4. 11.446. 6. 113.7234. Art. 243; pages 170 and 171. 2.1-2?/. 4. 4:x-3ab. 6. y 2 — y-1. 8. a; 2 -2a; + l. 3. 2ar + 3. 5. a: 2 + 2a;-4. 7. x + -. x 9. a + b + c. 10. 2 a; 2 3a;-l. 115 11. X+'„ -o-K— R + 3 a; 2 9 a; 6 ' 81a; 8 12. x ANSWERS TO EXAMPLES. 435 a? a G 5 a? g 2 1 1 5 3x 2 ~9x 5 Six 8 '" . 4z 4 32z 10 768x 16 "* Art. 245; page 173. 2. 123. 5. «. 8. 1.442. 11. .855. 3. .898. 6. 3.72. 9. 1.913. 12. .420.* 4. 11.4. 7. .0803. 10. 5.963. 13. .561. Art. 247; pages 175 and 176. 2. to 2 — 2m — 4. 3. a 2 -ax + x 2 . 4. 2x — 1. 5. x' 2 — x + 1. Art. 248 ; page 176. 1. 2x-3y. 2. cr-1. 3. to 2 -2to-3. Art. 257 ; pages 180 and 181. 4. c^. 5. af*. 6. *»*. 9. - 6 a J*K 11. a 4 6~ 4 - 2 + a- 4 i 4 . 12. a - b. 13. a- 5 -3a~ 3 b 2 + a- 2 b s -2a- 1 b i . 14. 18a 2 i 2 +10 + 2«- 2 &- 2 . 15. 2 a r 1 2/-10z Z /- 1 + 8.ry- 3 - 16. 2-4x"V + 2affy. 17. 6x 2 -7^-19af*+5a;+9^-2^. 18. 32aZr 2 -50+18 a - 1 & 2 Art. 258; pages 182 and 183. 5. <T*. 6. to 5 . 7. x 12 \ 8. ^ ' 11. J + Jb^ + Jb^ + Jb^ + bl 12. a-^^^ + i- 2 13. x- 3 f -x~ 2 y + 2x-\ 14. a>* y- 1 - 3 + 4 x~* y, 15. x- 1 y- 2 -a;- 2 7/- 3 -x- 3 y- 4 . 16. 2x^y~ ?J -x~^y-x~^ >*. Art. 260 ; page 184. 6. x*. 7. c~K 8. to- 1 . 9. 2T 3 . 10. a~ 3 . 11. w" 1 436 ALGEBRA. Art. 262; page 186. ' 1 n 12500 3 - 9 - 5 -ioooo- 7 - 4 ' 9± -^~ 4. ±216. 6. ±-^. 8.-243. 10. ±£. Art. 263 ; pages 186 and 187. 5. 3 3-2 y _ 2 x- 1 - 2T 1 - 6. 2 aj^ + x y~^ -4^ y~*. ./•' 7. a?t"y T *-2-+aT*y*. 1L 2y*-y*ar 1 . 12. 13. a;- 2 " 6 . 14. a*». 15. a 31 . 16. a - ^. 17. sc. 01 •* + «* oo h^d'-afd* 9q 5 »" (»»-!) 21 1_3« 3 * ^ ab*# + aP*' 3a ' Art. 267; page 189. 2. ^27, #16, #25. 3. $625, ^216, ^49. 12.^ 12, 12 „ 15,— — 15,-———- 15,. 4. V*V, V^ 4 ^ VVs 3 - 5. #32 a 5 , #27 & 3 , #64c 3 . 12. 12, 6. #cr+2a6+6 2 , \/a 3 -3a-b+3ab 2 -b 3 . 7. #a°-3a 4 .z 2 +3a 2 ;c 4 -x 6 , fa 8 -2« 3 x 3 + x 6 . 8. y^3. 9. #2. 10. #4. Art. 269 ; page 190. 5. \l$ab\ 6. #a& 2 . 7. v/(|f 3 )- Art. 270 ; page 191. 11. 3xyS/2xy' 2 -3xhj. 12. (a — 3)^a. 13. (x + y)^x — y. 14. (2 x + 3 a) f 6a. 15. 4 a & #3a& 2 + 5&. 18. ^ 6 - 1 ._ „„ 1 ,„. „. 2a 19. ^i/30. 20. Ji/21. 21. ^f y/3. 22. r> #6*. 6 b y J ANSWERS TO EXAMPLES. 437 a\j ab c 2 (a + x) 6- (a + 6) Art. 272; page 192. 7. v^x s- ft^F- 9- y/(^)- Art. 273; pages 193 and 194. «/. o 38 3. 10 si 2. 4. 12 v/ 3. 5. 9^2. 6. ^y/o 7- |\/6. 8. |j2 + ^18. 9.4^5. 10. i^- 11.6asJ3a. 12. ^V 3 - 13. ^2 + ^3. 14. 2 yV 2 - tf. 15. (2 a - 5 b) \Jl x. Art. 274; pages 195 and 196. 5. \f^¥x-\ 6.^4500^ 7 - ^(g^ga)- 8. y/5^ . 10. a; + v/z - 6. 11. 21 x - 38 v^ + 5. 12.2. 13.-1. U. x-ij-z + 2\jy^. 15.4 + 2^10. 16. 56+12v/35. 17.36-32^15. 18. ax-x\ 19. m + TO. 20. 14 — 4^6. 21. 147 + 30 v/24. 22. 1 + 2 a %/l-a 2 . 2-3. 2 a- 2 \/a 2 -& 2 . Art. 275 ; page 196. b/16 V / 243' e/8 V9- 7. {/" 8. {18. 9. ^ 438 ALGEBRA. Art. 276; page 197. 3. ^125. 4. y/7. 5. 2304 x\ 6. a 4 * 2 . 7. \j~a~^-~b. 8. SlaHx Sjbx. 9. a: 2 + 2x + l. 10. 16 a: 2 - 48. Art. 277; pages 198 and 199. 12,- 3.^2. 4. si 2. 5.<)a + b. 6. V* - 1. 7. y/2 8. ^3. 9. v/3. 10. ^^V- n - V 2 - Art. 278 ; page 200. . 3s/2 . ^4V 2 x 5 \/ 2 a 2°\/3tf O. pr — . 4. — s . 0. — ^ — . D. 5 2 2a 2 3a Art. 279; page 201. 3. 12 ~ 4y/2 . 4. 5 + 2 V 3. 5. 2 ^ 6 - 5. a + 2_^b'+b 16 + 7 y/ 10 6 ' " «-fi ^ ~* 7 * " 13 _ a — 2 ^ ax + x a + 3 + 3 \la + 1 a — x a a+^a*-x 2 10. 2a 2 -l-2aV / « 2 -l- H. • 12. y^-l-ff 2 . a; 1Q x*-2 + x^x 3 -4: 14 a: - 24- 11 Vs 8 - 2a; 2 18 — o x Art. 281; page 202. 2. .894. 3. 7.243. 4. 3.365. 5. .101. Art. 286; page 204. 4. — 8 v/6. 5. 12\/ab. 6. 46. 7. 2. 8. -abc V^l. 9. a 2 + b. 10. 12. 12. sJB. ANSWERS TO EXAMPLES. 439 13. y/2. . 14. y/5. 15. ^3. 16. \f^l. 17. 1 + <f^2. 18. 2 (f- b ) . 19. 1 _ 4 y/^3. 20. - 100 - 18 tf=2. or + b T Art. 293; pages 207 and 208. 5. v/T + y/5. 8. 5 + y/lO. 11.^15-^5.14.3-2^-2. 6. v/21-v/3. 9. 3-v/3. 12.3 + ^5. 15. 5 + 3 \/^2. 7. 3 + v/7. 10. v^5-v/3. 13. 7-3 y/ 2. 16. 6-V-l. 17. ^m + n— \jm — n. 18. x — SJax. 19. 3 + v/2. 20. v/2-1. 21. 2-^3. Art. 297; pages 209 and 210. 4. 17. 9. 4. 14. 4. 19. -1. 24. 4. 5. 19. 10. 5. 15. 81. 20. -3. 25. 5. 6. 7f. 11. -2. 16. 4. 21. 4. 26. 3. 7. 2. 12. |. 17. 8. 22. 12. 27. 6. 8. 4. 13. 4. 18. -3. 23. 25. 28. 39. 29. 3£. 30. 3. 31. 6. 32. 3«-l Art. 303 ; pages 212 and 213. 2. ±3. 4. ±y/ (_?). 6. ±7. 8. ±1. 10. ± 3. ±5. 5. ±1. 7. iy/11. 9. ±1. 11. ± Ifr — IA 12. ±^\L-±y 13. ±v/a + ft. Art. 310 ; pages 220 to 222. 10. 5 or - 7. 11. 11 or - 2. 12. 5 or 3. 440 ALGEBRA. 13. - 5 or - 13. 29. - 4 or - 1. 45. 2. 14. ^or-|. 30. 2 or i. 46. 4 or 0. 15. 2 or -. ' 31. 4 or - If. 47. 3 or - 2. 16,_l r-|. 32. 4±2y/3. 48. -2ov~ do bo 17. ^^^ 33.3or-l. 49. ± 2 12 --■ — ~ .y/8' 18. 17± y 337 . 34. 2 or - 1 . 50. 25 or 3. 4 7 19. -- or -1 35. 7 or | . 51. 6 or -2. o 2 6 20. lor- 7 , 36. 4or-^. 52. -or--. 4 4 a. c 21. =f or - 2. 37. - 10 ± v^78. 53. a ± b. o 0Q 1±V409 3 a a **. £ . 38. — 3i or — 2^. 54. — — - or - . b ""42 23. — or - . 39. 1 or — . 55. — a or — b. 4 Z oh 24. 3£ or - 1. 40. 1 or £ . 56. 11 or 18£. 25. 13 or - 2. 41. 5 or ^ . 57. 5 or - 3. 5 26. I or i . 42. 18 or 3. 58. 12± J . 2 14 5 27. 1 or 3i. 43. — 2 or — . 59. a — ft or — a — c. 28. - 4 or - ** . 44. - 3 or 2*. 60. ^=* or- 3 ^. _, a + b a — b bl. or — — r . a — o a + ANSWERS TO EXAMPLES. 441 Art. 311 ; pages 224 to 227. 4. 12 rds. 5. 40000 sq. rds., and 14400 sq. rds. 6. 9 and 6. 7. 16 and 10. 8. 16. 9. 3 inches. 10. $ 30. 11. 14 and 5. 12. $2000. 13. 18bbls., at $4 each. 14. 256 sq. yds. 15.5. 16. 7 and 8. 17. 7, 8, and 9. 18. Length, 125 ; breadth, 50. 19. 9. 20. 3712. 21. 80. 22. 20. 23. Area of court, 529 square yards ; width of walk, 4 yards. 24. 36 bu. at $1.40. 25. Larger, $77.17^; smaller, $56.70. 26. 1st, 14400; 2d, 625 ; or, 1st, 8464 ; 2d, 6561. 27. 84. 28. 6. 29. Larger pipe, 5 hours ; smaller, 7 hours. 30. 38 or 266 miles. 31. 70 miles. Art. 314 ; pages 230 to 232. 5. ±3or±V-13. 6. ±l or ±J_ 7. lor -2. I y 5 8. ± 1 or ± - . 9. ± 7 or ± 5. 10. ^3 or - ^23. 11. ± 8 or ± d(- ^ . 12. 4 or ^49. 13. 4 or 1. 14. 243 or - J' (28 5 ). 15. 4 or 1\. 16. 49 or 25. 18. 2, - 2, 3, or 7. 19. 3 or - 1. 20. ± 1 or ± 2. 21. 2 or - 3. 23. 1, - 1, 5, or 7. 24. 2, - 3, 4, or - 5. 25. 1, 2, - 5, or 8. 26. 1, - 1, - 6, or - 8. 28. 3,-|or 3± 4 V/ 55 . 29. 8, -2, or 3 ± y/ 110. 80.?,-?, or " 3± 2 2 ^ 3 . 31. 1,9, or 5±2 N /2. 32. 0,-5, 1, or-^. Art. 317; page 234. 2. x = 2 } y—±l; or, x = — 2, y=±l. 3. x = 4, y = ± 5 ; 442 ALGEBRA. „ . 1 1 11 or, x= — 4, y=±5. 4. « = g, y — ±^ ° r ' a;=— 3' ^ =± 2' 1 1 5. x = 3,y = ±p or,x = — 3,y = ±g. Art. 318; page 235. 2. a? = 7, y = — 8; or, x = — 8, y = 7. 3. a; = 5, y = — 2 ; or, a; = — 2, y = 5. 4. x = 3, y = 4 ; or, cc = — 4, y = — 3. 5 5 5. x = S, y = 2> or > cc=: ~2' y:= — 8 * 1 5 6. x = 2, y = 4 ; or,x = — ^,y = ^. 7. ar = 2, y = — 3 ; or, x = 3, y = — 2. 8. x = 1, y = 2 ; or, a; = 2, y = 1. Q Q 9 15 62 9. a? = 3, y — 2; or,x = — —,y = —. 10. a? = 9, y = 6 ; or, x = — 6, y = — 9. 11. a; = 2, y = 9 ; or, x = 9, y = 2. 12. a; = 9, y = 3 ; or, x = — 3, y = — 9. 13. x = 6, y = — 4 ; or, x = — 4, ?/ = 6. 14. a; = 3, y = 2; or, a; = —, y = — — . 15. x = 5, y = 3 ; or, a; = — 3, y = — 5. 16. x = 3, y = — 7 ; or, a; = — 7, y = 3. Art. 319; page 238. 4. x = 3, y = 4; a; = 4, y = 3; x = — 3, y = — 4; or, x = — 4, y = — 3. U. & = 6, y = 7 ; a; = 7, y = 6 ; a; = — 6, y = — 7; or, a; = — 7, y = — 6. 6. x = 2, y = — 3 ; or, x = — 3, y = 2. ANSWERS TO EXAMPLES. 443 7. x = — 1, y = 4 ; or, as = — 4, y == 1. 8. # =1 3, y = — 2 ; or, x = — 2, y = 3. 9. a; = 4, y = — 7 ; or, a; = 7, y = — 4. 10. x = 5, y = 6 ; or, cc — 6, y = 5. 11. x = 5, y = 2 ; or, a; = — 2, y = — 5. Art. 320 ; pages 239 and 240. 2. x = 2, y = 2> x = - 2 ,!/ = -2> x = \ 5>y = — 2 \ 2 5 ; 5 g 3. x = 2,y = 3; x = — 2,y = — S- x = ^^ r ,y = ^31 ' y ~ v' 31 ' 5 6 y/31' y "~^31.' 4. aj = 3, y = l; a>= — 3, y = — 1; x = 2^2, y = \/2; or, ^ = -2^2, y = — ^2. 5. £ = 3, y = 5;a; = — 3, y = — 5; x = -, y = y5 or, z = — -, 13 6. a; = 2, y = — l\ x = — 2,y = l-, x = —-r , y 5 7 7. a; = 2, y — 1 ; a; = — 2, ?/ = — 1 ; x = 7, y = — 19 ; or, x — — 7, y = 19. Art 321; pages 243 and 244. 5. x = l, y = 8; or, x = 8, y = 1. 6. a; = 4, y = 9 ; or, x = 9, y = 4. 7. a: = 2, ?/ = 3 ; or, a: = 3, y = 2. 444 ALGEBRA. 8. x=3, y=4 ; 3=4, y=3 ; x=-± + \J^11, y =-A-\^H ■ or, x = — 4 — \/— 11, y = — 4 + y^— 11. 9. a:=4,y=5;a:=16, y=-7 ; 3=- 12+^/58, y=-l— ^58 ; or, x =- 12 - y/58, y = - 1 + y/58. 10. a; = 4, y = 2 ; cc — — 2, y = — 4 ; or, a; = 0, y = 0. 11 o a 605 20 io -, 3 11. x = 9, y = 4; or, *=.— , y = — . 12. a- = 1, y = -. 13. x = 3, y = 2 ; or, a; = 2, y = 3. 14. x — 9, y = 4. 15. 3 = 1, y=-3; a=-3, y=l; a-=l+V^^2, y=l- V^2; or, a; = 1 — \/-2, y = 1 + ^~2. 1« 1 o o -, 3+V-55 -3+V-55. 16. x=l,y=—2;x=2,y=—l;x= 1 ,y= j ; or,x = , y = y. . 17 9 o o o -1+3V^3 1+3^. 17. x=2,y = 3;a:=-3,y=-2;a: = ^ ,y= J , -1-3 y/^3 1-3 V / ' =r 3 or, a;= ^ » ? = ^ ■ 18 . , = 3 ;y =2;, = 2 ,,=3 i ,=^±^,,==^. 9 ; _-9-v/309 -9 + ^309 12 " ,y ~ 12 19. 3=1, y=-3; 3=-l, y=3; 3=141, y=3f; or, 3=-14f, 20. 3 = 2,y = 3; or,3 = 2f>y = lf. 01,100 4 22 59 21. a: = 4, y = 2, « = 3; or, x = -,ij = — , z = -g-- 22. a; = 1, y = 2, z = 4 ; a; = — 1, y = — 2, s = — 4 ; x — 9, y = — 6, s = 4 ; or, x = — 9, y = 6, z = — 4. Art. 322; pages 246 to 248. 4. 12 and 7, or — 12 and — 7. 5. 11 and 7, or — 11 and — 7. 6. A, $2025; B, $900; or, A, $900; B, $2025. ANSWERS TO EXAMPLES. 445 7. A, 25 ; B, 30. 8. Length, 150 yds. ; breadth, 100 yds. 9. 13 and 6. 10. A, $15; B, $80. 11. 10 lbs., at 8 cts. 12. A, $ 5 ; B, $ 120. 13. Duck, $ 0.75 ; turkey, $ 1.25. 14. Price, $ 1600 ; length, 1G0 rods ; breadth, 40 rods. 15. Larger, 864 sq. in. ; smaller, 384. 16. A, $ 275 ; B, $ 225. 17. 1st rate, 7 p.c; 2d, 6. 18. A, 40 acres at $ 8 ; B, 64, at $ 5. 19. Distance of towns, 450 miles ; A, 30 miles a day ; B, 25. 20. 3 and 1 ; or, 2 + sj 7 and 2 - y/ 7. 21. Larger, 12 ft. ; smaller, 9. 22. Width of street, 63 ft.; length of ladder, 45. 23. B, 15 days ; C, 18 days. 24. Length, 16 yds. ; width, 2 yds. Art. 328 ; page 253. 3. (a: + 60) (a; + 13). 6. (x + 13) (a? -3). 9. (4a;-l) (2a; + 5). 4. (x -9) (x -2). 7. (jc-5)(2a; + 3). 10. (x - 3) (4 x - 3). 5. (a; -10) (a; + 6). 8. (7 a; + 3) (3 a; + 7). 11. (a; + 2) (2 a; -3). 12. (3a:-2 + v/3)(3a-^2- v /3). 13. (y/17 + 4 + a-) (^17-4-3:). 14. (7a3 + l + 2 v / 5)(7a; + l-2v/5). Art. 329 ; page 254. 2. ar + a; = 2. 5. 3 x 2 - 2 x = 133. 8. 3 x 1 + 17 x = 0. 3. a; 2 -9 a; =-20. 6. 21.x 2 + 44 x = 32. 9. a: 2 -2a; = 4. 4. 5a; 2 - 12a; = 9. 7. 6ar+35a;=-49. 10. Art. 330 ; page 255. x 2 — 2 m x=n— m 4 7. 0or y . 8. or — 4. 9. or ± 3. in 5 x lL-*orl a c 12. ± 2 or ± 3. 1 5 13 --3° r± 2 14. ±sja or — . 15. 0, - 5 7 1 ■2'3'° r ~4- 16. 2,3,-3, -4, I, or -5. 446 ALGEBRA. Art. 331; page 256. 3. (x + y/2 x + 1) (x - \/2 x + 1). 4. (a; + \Jx + 1) (x — \J x + 1). 5. (a + ^5lTb + b) (a - \j~h~ab + b). 6. (x°~ + 3 x y + f) (x 2 -3xy + f). 7. (x + 1 + S/Sx + 2) (x + 1 - \J~3x~+2). 8. (m 2 + mn + w 2 ) (m 2 — m n + to 2 ). Art. 332 ; page 256. 2. 7. ^ 7 * V 71 - 1 ™ -v / 7±V /: -l or 2^2 2^2 Art. 357; pages 269 and 270. 1. 4. 2. 11. 3. £. 4. 1 J. 5. ±4. 6. ± 12. 7. ± 14. 8. 25 and 20. 9. 23 and 27. 10. 12 and 15. 11. 8 and 18. 12. 26 and 14. 13. 17 and 12. 14. 12 and 8. 15. First, 1:2; second, 2 : 1. 16. Females : males = 4:5. 17. 8 : 7. Art. 365; page 273. 2.4. 3. „ = 8* 4* 5.4. 6. y= 14 . . °> 4 — 5cc 7. 10 inches. 8. 3 (y/ 2 — 1) inches. 9.143. ANSWERS TO EXAMPLES. 447 Art. 370; page 276. 3. 1=71, £=540. 4. Z=-69, £=-620. 5. 1=57, £=552. OQ A9 3 6. l=-U5, £=-2175. 7. Z=^|,£=y. 8. l=--,S=0. 9. l=-^,S=±. 10. l=~,S=~. U.l=5,S=17. 4 2 Art. 371; pages 278 and 279. q 95 1 4. a = 3,£ = 741. 5. a = ^,i==--^-. 6. <Z=-,£=39. 7. d = — i, « = -?. 8. a=o,d=-3. 9. w=18, £=411. 12 4 10. «Z = — 8, n = ll. 11. w = 30, Z = 80. 12. ^ = 52, a = 4; or, » = 43, a — — 5. 13. n = l%l = - 43. Art. 372; page 279. 7 8 10 11 5 3 1 1 A 3'3"I'"3"" J, 2' w, 2' '2' ' 2' 4 2 3 _ 4 _ 5 5 Jl -^ _^ _!? -^ -?? *■ ^ — ^ 4 > °- 7 ' 7' 7' 7' 7' 7 2 6 14 22 „ am + 5 a(m -l) + 2b 5'5' 5 ' 5 ' ' m + 1 ' m + 1 Art. 373; page 281. 3. 2500. 4. Last payment, $ 103 ; amount, $ 2704. 5. 4. 6. After 9 days, at a distance of 90 leagues. 7. 4, 11, 18, and 25. 8. 3. 9. 0. 10. 20 miles. 11. 2, 6, 10, and 14 ; or, - 2, - 6, - 10, and - 14. 12. 8. 448 ALGEBRA. Art. 378; page 284. 4. 1 = 2048, £=4095. 9. 1 = - — , s = -^^ 64 ' 192 64 2059 _ 1 511 5- l = m> S= 243-. 10. 1 = ^, S=~. 6. Z = 2048, £=1638. 11. l^-~, £=^. 7 l--- L s- 3 ^ 12 Z- 1 <?- 341 *~ 256' ^-256' "' ^-~768'^-~256- 1 2047 8 - l = zm> S =mi- 13^ = 192,5=129. Art. 379; page 286. - 1 c 341 2 _ 2 4. a = ^, £ = 7 r-. 5. a = 7., Z = 2' 2 "" "- _ 3' 6561' 6. r = 3, £=2186; or, r = -3, £=1094. 1 2457 7. ^-j, /S= m . 8. » = 5, £=121. 9. n = l, r = \. 10. w = 6, Z = -?S 1L w = 8, a = -l. ^ 2 Art. 380 ; pages 287 and 288. 3. 4. 5 3 160 y - 19" 4 ? 3 ft 15 Art. 381 8 'i' ; page 288. ,0.-12. 3.|. 4. 13 27" 5 n 5 - 15- ft 8G 7 6 - 165* 7 ' 17 237 150* 1100 ANSWERS TO EXAMPLES. 449 Art. 382; pages 289 and 290. 48 16 32 64 392781243 39 '3' 9' 27' 81' 243" 2' 2' 2 ' 2 ' 2 ' ° r ' 2' 2' -|, |, -y. 5. - 6, -18, - 54, - 162, - 486, - 1458. 927 81243 33333 3 3 ~4' 16' ""64' 256* 4' 8' 16' 32' 64' 128' 256' 3 3 _3_ _3^ 3_ _3_ _3_ ° r '~4' 8' ~16' 32' 64' 128' " 256' Art. 383; page 291. 3. $ 64. 4. $ 295.23. 5. 3100 ft. 6. 5, 10, 20, and 40 ; or, - 15, 30, - 60, and 120. 7. - 4. 8. T V Art. 386 ; page 292. Z .E. 3. -i. 4. 3 " •» 31' 78' ' 4' ' arc-&?i + 2&-a' 2. Art. 387; page 293. 48 24 16 12 48 8 48 125' 65' 45' 35' 145' 25' 155' 5_5_5 ■ 21 _7 21 21 3 ' 4'" 3'" 2" '" '" 5 ' 3' 13' 17" _ (m + 1) a 6 (m + 1) a b (m + 1) ab m b + a ' m6 + 2a-6' mH3«-2i' Art. 397; pages 297 and 298. 4. Of 4 letters, 360 ; of 3, 120 ; of 6, 720 ; in all, 1956. 5. 1680. 6. 3838380. 7. 358800. 8. 15120. 9. 120. 10. 35. 11. 15504. 12. 31824. 13. 77520. 14. 648. 450 ALGEBRA. Art. 403 ; page 302. 5. 1 + 5 c + 10 c 2 + 10 c 3 + 5 c 4 + c 5 . 6. a 6 + 6 a 5 a 8 + 15 a 4 x 6 + 20 a 3 x 9 + 15 a?x u +Qa x lh + x ls . 7. x s -8x 6 y + 24:X i 7f-S2x 2 f+16y\ 8. a 7 ^-7a 6 i 6 c^ + 21a 5 i 5 cV 2 -35aH 4 c 8 ^ 3 +35aH 8 c 4 ^ 4 -21a 2 & 2 c 5 d 5 +7a&c 6 a ,6 -c 7 cf. 9. m 12 + 18 m 10 w 2 + 135 m 8 w 4 + 540 m 6 w 6 + 1215 m 4 w 8 + 1458 m- n 10 + 729 w 12 . 10. a - 10 -20 a- 8 a^+160 ar 6 x- 640a~ ^+1280 or 2 ar-1024a^. 11. c™ + 8 c*' cfl + 28 c 4 e^ + 56 c$~ <fi + 70 c% d 3 + 56 c 2 d^' + 28 c* <# + 8 J eft + d 6 . 12. m~^ + 14»"^ w 3 + 84 m- 8 » 6 +280 m - ^" w 9 + 560m~*» 12 + 672 m"^ w 15 + 448 j»~* w" 4- 128 w 21 . 13. a- 4 -4ffl- 3 i 2 x^ + 6 a~ 2 Z> 4 a;^ - 4 or 1 6 6 a; + b s sA Art. 404; page 303. 2. 5005 a 6 a; 9 . 4. - 19448 c 10 d 7 . 6. 42240 x~ 3 yK 3.2002 m 6 . 5.495 a 8 . 7. 262440 a 2 ar 7 . Art. 405 ; page 304. 2. 1 - 4 x + 2 a; 2 + 8 x 3 - 5 x i - 8 x h + 2 x 6 + 4 x 1 + x\ 3. a; 6 + 9 x 5 + 30 a; 4 + 45 x 3 + 30 x 2 + 9 x + 1. 4. l-6a; + 6a; 2 +16a: 8 — 12a 4 -24a; 8 -8a: 6 . 5. l+5a; + 5a; 2 -10a; 3 -15a; 4 +lla; 5 +15x 6 -10a; 7 -5x 8 +5x 9 -cc 10 . Art. 414; page 309. 3. l-2a: + 2a; 2 -2a: 3 + 2a; 4 4. 3 + 19 x + 95 x 1 + 475 x a + 2375 a; 4 5. 2-a; + 3a: 2 -a: 8 + 3a: 4 ANSWERS TO EXAMPLES. 451 6. l-2a; + 2a; 3 — 2x 4 + 2x 6 7. l-2x + ox 2 -16x 3 + 4:7x i 1 5x 7_x* lTjc 3 31_x 4 9. 2- 7 a; + 28a; 2 - 91 a; 3 + 322 a; 4 2 a; 7 a; 2 13 a; 3 8 a; 4 io. i + — -— 27 + 81 1 3jc a 2 15k 8 49a 4 1L 2 + ~T + IT* 16 + 32 Art. 415; page 310. 2 a;- 2 4 a;- 1 _8_ 16 a; 32 a; 2 2 - _ 3 _+_ 9 _+ 27 + 81 + 243 3, £C - 1 + 3 + 2x-5a; 2 -16a; 3 4. x~ 2 — x- 1 -2 x + 2 a; 2 -4 a; 3 Art. 416; page 311. a; a; 2 a; 3 5 a; 4 a; 3 a; 2 3 X s 3 a; 4 3 * 1 + 2~8~ + 16~128"" ,1+ 2 + 8 "16 + 128"' a; 2 a; 3 5 a; 4 _ a _ s 2 5 a; 3 10 a; 4 3. l_aj_— — - — g-... «>. i 3 9 -~ 81 - 243 •'• x 4 „ . a; 2 a: 2 13 a; 3 8 a; 4 4. l_a; + a; 2 + a; 3 + y ... 7. 1 + -+— — ^ +-^3 ... Art. 418 ; page 314. 3 2.41 _J_ 6_ 2 - x~T2 + x~=2' x-2 x 'x-7 x-6' , 3 2 ,1 J_ 7 _2 3_ «~T+8" & '^T4 + a; + l' '"2»-5 Sas + l" Q 1 2 _J 1 1 4 8 "3+T£ + 3=!T y "6(x + l) 2(a;-l) i "3(a;-2)' 452 ALGEBRA. Art. 419 ; page 316. o 11 1 „ 1 4 4 2- — ^- + 7— ^Tv 2 + 7— "TV,- 4.-— + - — + x + l (x + l) 2 (x + l) 3 ' x-2 (x-2f ' (a; -2) 3 2 3 5 3 6 B' x-5 (x-5) 2 ' x + l (x + l)' 2 (x + l) 3 ' 3 5 6. 7. 2 (2 a; - 5) 2 (2 x - 5) 2 ' 2 4 3 3 a; + 2 (3x + 2) 2 (3x + 2) 3 Art. 420 ; page 317. _ 2 3 5 .515 2. — -, — ; . 4. x x+2 (x+2) 2 ' 'x x* x + l (x + l) 2 ' „11 1 1 .123 4 3. -+— T + —0 + 7— 2K3- 5.- ,+ x x— 1 x—2 (x—2)' 2 ' ' x x 2 x s x + 5* ■i x-2 2x-3 (2x-3) „ 5 1 2 5 4 7. S + -T-— X X 2 X 3 X + l (x+l) 2 ' Art. 422; page 320. o 2,84 /i y 3?/ 19?/ 6 19 y' 3. x = y — ?/ 2 + ?/ 8 — y 4 . . . 4. x = sL -j £ — . . J J J J 2 16 + 128 128 5. x = y + y 3 + 2 y 5 + 5 y 7 . . . vy ; 2 i- 3 4 ■" 7 x _ ?/ , ^,2^ 17 y' y 2y 2 y 3 14y* ^~ y+ 3 + l5- + 315"- 8 ' *-3 + 27~~243~2T87" Art. 425; page 325. - § 5 | .15 1,5 a . 5 _a 4. a 2 + -a 2 x + -^-a 2 x' + z—a 2 x z — — -a ^ 4 «s 8 lb 128 ANSWERS TO EXAMPLES. 453 5. 1 - 6 x + 21 x 2 - 56 x 3 + 126 x* 3 12 2 52 3 234 4 b. i + 5 x + 25 x + 125 x + G25 x i 1 _i. 1 _| 1 _i 5 -z 7. a J —-^a *x — -^a 'X' — zr^a 2 x 6 — zr^a J x* £ o lb lwo „ , 1 2 „ 14, 35 8.1-- x + - x *-- x , + mX > 9. ar 3 + 3a- 4 x + 6 a~ 5 x 2 + 10 a~ 6 x 3 + 15 «" 7 x* .... 10. c"^— c- 3 df + c~* d 2 - c- 6 d 3 + c" 1 *' d* 11. x z —2x b y — X s y 2 — -X s y 3 — ^x 3 y* o o I 3 15 7 35 3 a 315 jjl 6 14. m + 6 in 6 n- + -jr- m J n 6 + -=- mr n 2 -\ — — m J ?i . . . 2 2 o in -i -i^ i ™ o o 1760 „ „ 12320 13. 1 — 10 xy- 1 + 80 a; 2 ?/- 2 q-b'jt'H 5— a? 4 ?/" 4 2 2 8 15. a 4 + 12 a 5 gT 2 + 90 a 6 y~ 4 + 540 a 7 jr 6 + 2835 a s y — 8 Art. 426 ; page 326. 33 aT'^'x'' 315 a 8 44 x^' y* "~2048 ' 128"' 6561 ' 663 x "V 4. 84 m 6 . 6. - " y ■ 8. 210 w^c" 8 . 8192 9. _?5? a -^aj- 6 . 10. 3§x- 3 «y- Vi z~ 1 £. Art. 427; page 327. 3. 3.14138. 5. 9.94987. 7. 2.03054. 4. 2.08008. 6. 1.96101. 8. 2.97183. 454 ALGEBRA. 2. 3. 1 + x Art. 435; pages 331 and 332. 4 -11 x „ 2 + 5« + 5a 2 1 — x — x 2 ' a b + ex 4. 5. 1 — 5 x + 6 x 2 1+x 1 - 2 x + x 2 ' 6. 7. (1 + a) 3 ' 3 — x — 6 x 2 l-2a;-a: 2 +2a; 3 3. 3. 8. 225. 8. l + 2cc 1 — x — x 2 9. 2 + 2 a; - 3 x 2 l — x + x 2 — x & 4. Art. 440; page 336. 14. 5. 30. 6. 1365. 7. 5050. 9. n i + 2n z + n 2 11. 165. 10. 6 n 6 + 15 m 4 + 10 w 8 ?« 30 12. 5525. 3. 4.0514. 1. 1.681241. 2. 2.644438. 3. 1.748188. Art. 443 ; pages 338 and 339. 4. 3.634241. 5. 2.23830. 6. 44.24. 7. $1,356. Art. 455 ; page 344. 4. 1.991226. 7. 2.225309. 5. 1.924279. 8. 3.848558. 6. 2.753582. 9. 2.702430. 10. 3.489536. 11. 4.191785. 12. 4158543. 1. 1.176091. 2. 2.096910. 3. 0.154902. Art. 456 ; page 345. 4. 2.243038. 7. 0.853872. 5. 0.522879. 8. 1.066947. 6. 1.045758. 9. 0.735954. Art 464; page 350. 2. 8.724276-10. 4. 9.470704-10. 6. 1.527511. 3. 1.714330. 5. 0.011739. 7. 8 780210-10. ANSWERS TO EXAMPLES. 4jj 8. 4.812917. 11. 9.942550-10. 14. 4.89381. 9. 7.013150-10. 12. 3 863506. 15. 1.718451. 10. 2.960116. 13. 8 640409-10. 16. 7.4984240-10. 17. 9.275374-10. 18. 1.9792784. 2. 76. 3. .2954. 4. 6.61005. 5. 55606.5. 6. .011089. Art. 465 ; page 352. 7. 186 334. 12. .034277. 8. .223905. 9. 1000.06. 10. 9.77667. • 11. .00130514. 17. .00548803. 18. 13. 46.7929. 14. 11.327. 15. 8.63076. 16. .2070207. 734.9114. Art. 466; pages 353 and 354. 1. 2.125240. 4. 3 108462. 7. 9.613158 - 10. 2. 8.223962-10. 5. 9.594161-10. 8. 9.970036-10. 3. 9.852169-10. 6. 7.315321-10. 9. 9.905232-10. Art. 468 ; pages 356 1 to 358. 1. .0341657. 13. 1.70869. 25. .580799. 2. .650573. 14. .788547. 26. -.631188. 3. 13560.2. 15. .680192. 27. 83.5656. 4. .136085. 16. 2.24328. 28. .297812. 5. 1.14720. • 17. .296850. 29. 98.4295. 6. 1.41421. 18. -.191680. 30. 1.65900. 7. 1.49535. 19. .644849. 31. 3 07616. 8. .0655264. 20. .501126. 32. .867674. 9. -1.97221. 21. 1.09872. 33. -2.09389 10. 458.623. 22. 1.06178. 34. 46809.2. 11. -.000113607. 23. 1.09328. 35. .588142. 12. 5.88336. 24. 1.65601. 36. 1.80446. 37. .00323011. 38. .0334343. 456 ALGEBRA. The following are the values of the expressions in Art. 468, when calculated by seven-figure logarithms : 1. .034165G8. 13. 1.708689. 25. .58079S7. 2. .0505727. 14. .7885469. 26. - .6311888 3. 13560 27. 15. .6801947. 27. 83.56558. 4. .1360851. 16. 2 243284. 28. .2978123. 5. 1.147203. 17. .2968501. 29. 98.42-J91. 6. 1.414214. 18. - .1916795. 30. 1.658989. 7. 1 495349. 19. .6443490. 31. 3.076162. 8. .06552632. 20. .5011282. 32. .8676754. 9. - 1.972211. 21. 1.098718. 33. - 2.093891. 10. 458.5759. 22. 1.061780. 34. 46808.95. 11. - .0001136063 . 23. 1.093280. 35. .5881412. 12. 5 883366. 24. 1.656005. 36. 1.804459. 37. .003230121 38. ,03343431. Art. 469 ; page 359 3. , 458156. 5. - .494903. 7. - -2.70951. 4. . 185339. 6. - .260231. 8. - - 10.2341. The results with seven-figure logarithms are as follows • 3. .4581568. 5. - .4949028. 7. - 2.709513. 4. .1853394. 6. -.2602272. 8. -10.23414. Art. 479 ; pages 368 and 369. 1. 7. 3. - 6. 5. 7. 7. 6. 2. 6. 4 .4 e. 5. 8. 7. 9. 1.56937. 13. 11.725 yrs. 17. 3.96913. 10. 2.44958. 14. $9756.59. 18. 7.18923. 11. 2.00906. 15. 7 per cent. 19. - 2.4578. 12. $5421.33. 16. 9.392 yrs. 20. - 1.07009 ANSWERS TO EXAMPLES. 457 The results of the last 12 examples, using seven-figure logarithms, are as follows : 9. 1.569369. 13. 11.725 yrs. 17. 3.969124. 10. 2.449576. 14. $9756.59. 18. 7.18922. 11. 2.009056. 15. 7 per cent, 19. -2.457802. 12. $5421.35. 16. 9.392 yrs. 20. -1.070092. Art. 489; page 373. 2. 3 and - 5. 3. a and | (- 1 ± y/^3). 4. 2 and 2. 5. ± 4. 6 . X 3_ 6a .2_ 6a ,_ 3 = 7> ? and 5 8 . ?and-l 3 2 4 5 Art. 490; page 374. 2. z 3 + 9x 2 + 23£ + 15 = 0. 4. 6ai 3 -lla: 2 + 6*-l = 0. 3. a 8 -19 a: -30 = 0. 5. cc 4 - 5x' + 4 = 0. 6. a; 4 - 10 x* + 35 x 2 - 50 cc + 24 = 0. 7. cc 3 - 13 x 2 + 56 x - 80 = 0. 8. x 4 -6a; 3 +5x 2 + 12x = 0. 9. 12 z 4 + 55 r 3 - 68 x 2 - 185 x + 150 = 0. Art. 494; page 375. 5 1. Sum, ; product, - 6. 2. Sum, - ; product, 12. 3. 2±2v/2. Art. 504; page 382. 2. ?/ 3 + 24?/ 2 + 191?/ + 498 = 0. 3. y* - 6?/ - if + 55y - 76 = 0. P ■: Art. 505; page 383. 2. y 2 -^ + <7 = 0. 4. ^-15y+26 = 0. y 3 + "2y ==0, 5 * 2/ 4 -6y 2 -137/-9 = 0. 458 ALGEBRA. Art. 513; page 388. 2. 1, 1, and 6. 4.-1,-1,-1, and 3. 3. 2, 2, and 3. 5. 2, 2, 2, and - 6. Art. 517; page 390. 2. - 1, 1, and 5. 3. 3. 4. 1. 5. 2. Art. 520; page 392. 3. 1 + s/U. 4. 1 + V 15 - 5- - (1 + V^)- 6. - (1 + $5). Art. 527; page 399. 3. Three ; respectively between and 1, 1 and 2, and — 1 and — 2. 4. Three ; two between 1 and 2, and one between — 3 and — 4. 5. One ; between 2 and 3. 6. Four ; respectively between and 1, 1 and 2, 2 and 3, and — 2 and — 3. 7. None. 8. Two ; respectively between 2 and 3, and 3 and 4. Art. 532; page 403. 3. — 1, - 2, and - 3. 9. A, and 1 ± }f—i. 4. 2, -2, and -3. 5. 2, 4, and - 1 ± y/^3. 6. - , 4, and — ^ • 7. 2, and — — *- . 8. 3. 6, and - 2. 10. 1, 2, and 3. 11. 3 9 , and ± 2 v/- -2. 12. o 13. 3. 14. 3, 4. — 3, and ■ — 5. ANSWERS TO EXAMPLES. 459 Art. 538; pages 407 and 408. 2 1 9 ±x /77 3 ± y/5 l_^ ±V /^_ 2j? -3 3. -1,1,1, or =^?. 6. 2,i,-3,or-|. 4. ±l J±V /=T,or^f: 3 . 7. l,5,i or2±^3. y/33- 5 ±^42- 10 y/33 - y/33 - 5 ± y/42 + 10 y/33 8 - 4 ' ° r ~I~ q _i 1 + y/ 5 ±V /2 V /5jri ~ Q " or 1-^5+^-2^5-10 a> *' 4 4 -l-^5±y^5-1 -l +v /5±y/-2v/5-10 iu. w, 2~ ? 01 o Art. 541; page 410. 3. -3or^±£3. 7. lorl±YE5. 2 2_ 4. 4 or 1 ± 4 \/=3. 8. 3 or 1 ± ^~ 3 ' 5. 3, 3, or -2. 9. 2, 2, or— 1. 6. 1,1, or -11. 10. $±-1(2. Art. 550; page 417. 2.2.09455. 3.7.61728. 4. 1.3569, 1.6920, and -3.0489. 5. 14.95407. 6. 2.2674 and 36796. 7. 2.85808,-60602, .44328, and - 3.90738. Art, 551; page 419. 2. 3.864854. 4. 2.4257. 6. 10.2609. 3. 4.11799. 5. .66437. 7. 8.414455. Art. 552; page 420. 2. 153209. 3. 1.02804. T A B L E CONTAINING THE LOGARITHMS OF NUMBERS FROM 1 TO 10,000. No. Log. No. Log. No. Lo£. No. Log. No. Loft. 1 0.000000 21 1.322219 41 1.612784 61 1.785330 81 1.908485 2 0.30103C 22 1.342423 42 1.623249 62 1.792392 82 1.913814 3 0.477121 23 1.361728 43 1.633468 63 1.799341 83 1.919078 4 0. CO 20 GO 24 1.380211 44 1.643453 64 1.806180 84 1.924279 6 0.698970 25 1.397940 45 1.653213 65 1.812913 85 1.929419 6 0.778151 26 1.414973 46 1.662758 66 1.819544 86 1.934498 7 0.845098 27 1.431364 47 1.672098 67 1.826075 87 1.939519 8 0.903090 28 1.447158 48 1.681241 68 i 1.832509 88 1.944483 9 0.954243 29 1.462398 49 1.690196 69 1.838849 89 1.949390 10 1.000000 30 1.477121 50 1.698970 70 1.845098 90 1.954243 11 1.041393 31 1.491362 51 1.707570 71 1.851258 91 1.959041 12 1.079181 32 1.505150 62 1.716003 72 1.857332 92 1.963788 13 1.113943 33 1.518514 53 1.724276 73 1.863323 93 1 .968483 14 1.146128 34 1.531479 54 1.732394 74 1.869232 94 1.973128 15 1.17C091 35 1.544068 55 1.740363 75 76 1.875061 95 1.977724 16 1.204120 36 1.556303 56 1.748188 1.880814 96 1.982271 17 1.230449 37 1.568202 57 1.755875 77 1.88641)1 97 1.986772 18 1.255273 38 1.579784 58 1.763428 78 1.892095 98 1.991226 19 I 1.278754 39 1.591065 59 1.770852 79 1.897627 99 1.995635 20 | 1.301030 40 1 .602060 60 1.778151 80 1.903090 too 2.000000 LOGARITHMS I 2 | 3 | 8 | 9 N.| I 1 D. 100,001)000 4321 8000 012837 7033 021189 6300 9384 0334 21 7420 000434 4751 9020 013259 7451 021003 6715 9789 033820 7825 110 1 2 3 4 6 7 S 9 041393 6323 9218 053078, 0905 0G0098 4458 8180 071882 6547 041787 5714 9000 053403 7280 001075 4832 8557 072250 5912 000808 6181 9451 013080 7808 022010 0125 030195 4227 8223 001301 5009 9876 014100 8284 022428 0533 030000 4028 8020 001734 6038 010300 4521 87CT, 02284 1 0942 031004 5029 9017 042182 0105 9993 053840 7066 001452 6200 8928 072017 0270 042570 6495 050380 4230 8040 061829 5580 9298 072985 6640 042909 0885 0507C6 , 4013 8426 002200 5953 9008 073352 7004 002100 6406 010724 4940 9110 023252 7350 031408 5430 9414 002598 6894 011147 5300 9532 023004 7757 031812 6830 9811 003029 7321 011570 5779 9947 024075 8164 032216 6230 040207 003401 7748 011993 6197 020301 4480 8571 032619 6629 040002 003891 432 8174428 012415'424 00l0i420 020775416 4896 412 8978408 033021 7028 040998 404 400 397 120 1 2 3 079181 082785 6360 9905 4 093422 5, 6910 6 100371 7- 3S04 8' 7210 91 10590 079543 083144 6710 090258 3772 7257 100715 4146 7549 110926 079904 083503 7071 09001 i 4122 7004 101059 4487 7888 111203 080206 3801 7420 090903 4471 7951 101403 4828 8227 111599 0801.20 080987 4219 4576 7781 8136 091315 091007 4820 6109 8298 8644 101747 102091 6109 5510 8505 8903 111934 112270 1149441 8265 121560! 4830J 8076J 131298 4496J 7071 140822! 3951 043302 7275 f 51153 4990 8805 002582 6326 070038 3718 7308 043755 7664 051538 5378 9185 062958 6699 070407 4085 7731 044148 8053 051924 6760 9563 003333 7071 070770 4451 8094 081347 4934 8490 092018 5518 8990 102434 6851 9241 112005 044540 8442 052309 0142 9942 063709 7443 071145 4816 8457 081707 6291 8845 092370 6866 9335 102777 6191 9579 112340 082007 5047 9198 092721 6215 9681 103119 6531 9916 113275 044932393 88301390 052694 6524 060320 4083 7815 0716141370 5182366 8819363 380 383 379 376 373 0824201360 6004 9552 093071 6502 100020 3462 6871 110253 3 CO 9 357 355 352 349 346 343 341 338 335 130,113943 1 7271 2 120574 3 3852 4 7105 5 130334 6 3539 7 6721 8 9879 9 143015 114277 7003 120903 4178 7429 130055 3858 7037 140194 3327 114011 7934 121231 4504 7753 130977 4177 7354 140508 3039 115278 8595 121888 5156 8399 131019 4814 7987 141136 4203 115011 8926 122216 5481 8722 131939 5133 8303 141450 4574 115943 9256 122544 5800 9045 132200 5451 8018 141703 4885 116276 9586 122871 6131 9308 132580 5769 8934 142070 5190 116608 9915 123198 6456 9090 132900 6080 9249 142389 5507 116940333 1-20245 330 3525 328 6781325 130012 323 3219 321 6403i318 9504 142702 5818 310 314 311 140 11 2 3 4 5 6 7 8 9 150 1 2 3 4 5 6 7 8 9 NTT 140128 9219 152288 6330 8302 101308 4353 7317 170202 3180 176091 8977 181844 4091 7521 190332 3125 5900 8057 201397 146438 9527 152594 5640 8664 101007 4050 7013 170555 3478 176381 9204 182129 4975 7803 190012 3403 0170 8932 201070 i 140748 9835 152900 5.U3 8005 101907 4947 7908 170848 3709 ,147058 150142 3205 0240 9206 1 02200 5244 8203 171141 4000, 170070 9552 182415 62591 8084 1 190892 31.81 0153 9206 201913 176959 9839 182700 6542 8306 191171 3959 0729 9481 202216 3 I 147307 150449 3510 0549 9507 102504 5541 8497 171434 4351 177248 180120 2985 6825 8047 191451 4 237 7005 9755 2(1 2 1 88 147070 'S0750 3815 3852 9808 102803 5838 8792 171720 4641 177530 180413 3270 0108 8928 191730 4514 7281 200029 2761 II 147985 151003 4120 7154 160108 3101 6134 9086 172019 4H32 177825 180099 3555 639J 9209 192010 4792 7556 200303 3033 o 148291 151370 4424 7457 100409 3400 6430 9380 172311 5222 178113 1801180 3839 6674 9490 19228H 6069 7832 1200577 3305 148003 151670 4728 7759 100709 3758 0720 9074 172003 6512 148911 151982 5032 f-001 161008 4055 7022 9908 172895 6802 309 307 305 303, 30ll 299- 297i 295 293 291 178401 181272 4123 6956 9771 192567 6346 8107 200850 3577 178689 289 181558 287 4407 285 7239.283 190051 '281 2840 279 8623 278 8382276 201124 271 3818 272 T~Ti | n OF NUMBERS. N.| | ] 2 3 4 5 6 7 8 9 ,D. 160 204120 204391 204663 204934 205204 205475 205746 206016 206286 206556)271 1 G82G 709G 73G5 7G34 7904 8173 8441 8710 8979 9247 269 2 9515 9783 210051 210319 210586 210853 211121 211388 211654 211921'2C7 3 212188 212454 2720 2986 3252 3518 3783 4049 4314 4579 266 4 4844 6109 6373 6638 5902 6166 6430 6694 6957 7221 264 5 7484 7747 8010 8273 8536 8798 9060 9323 9585 9846 262 6 220108 220370 220631 220892 221153 221414 221675 221936 222196 222456 261 7 2716 297G 3236 3496 3755 4015 4274 4533 4792 6051 259 8 5309 5568 5826 6084 6342 6600 6858 7115 7372 7630 258 9 7887 8144 8400 8057 231215 8913 231470 9170 231724 9426 23T9~79 9G82 232234 9938 230193 256 170 230449 230704 230960 232488 232742 255 1 2996 3250 3504 3757 4011 4264 4517 4770 6023 6276 253 2 5528 5781 6033 6285 6537 6789 7041 7292 7544 7795 252 3 804G 1 8297 8548 8799 9049 9299 9550 9800 240050 240300 250 4 240549 240799 241048 241297 241546 241795 242044 242293 2541 2790 249 6 3038 3286 3534 3782 4030 4277 4525 4772 5019 6266 248 6 5513 5759 6006 6252 6499 6745 6991 7237 7482 7728 246 7 7973 8219 84G4 8709 8954 9198 9443 9687 9932 250176 •245 8 250420 250GG4 250908 251151 251395 251638 251881 252125 252368 2610 24o 9 2853 3096 255514 3338 255755 3580| 3822 4064 4306 256718 4548 4790 5031 242 180 255273 255996 256237 256477 256958 257198 257439 241 1 7679 7918 8158 8398 8637 8877 9116 9355 9594 9833 239 2 260071 260310 260548 260787 261025 261263 261501 261739 261976 262214 238 3 2451 2G88 2925 3162 3399 3G3G 3873 4109 4346 4582 237 4 4818 5054 6290 5525 6761 6996 6232 6467 6702 G037 235 6 7172 7406 7641 7875 8110 8344 8578 8812 904G 9279 234 6 9513 9746 9980 270213 270446 270679 270912 271144 271377 271609 233 1 271842 272074 272306 2538 2770 3001 3233 3464 3696 3927 232 8 4158 4389 4620 4850 5081 5311 6542 6772 6002 6232 230 9 190 6462 278754 6692 6921 7151 7380 7609 7838 8067 8296 8525 229 278982:279211 2794391279667, 281715281942 279895 28 280351 21 228 1 281033 281261:281488 282169 2396 2622 2849 3075 227 2 3301 3527 3753 3979 4205 4431 4656 4882 5107 5332 226 3 6557 5782! G007 6232 6456 6681 6905 7130 7354 7578,225 4 7802 8026 8249 8473 8696 8920 9143 93G6 9589 9812 223 5 290035 2902571290480 290702 290925 291147 291369 291591 291813 292034 222 6 2256 2478 2699 2920 3141 3363 3584 3804 4025 4246 221 7 4466 4687 4907 5127 5347 5567 5787 6007 6226 6446 220 8 6665 6884 7104 7323 7542 7761 7979 8198 8416 8G35 219 9 8853 9071 9289 9507 9725 9943 3001G1 300378 300595 300813 218 200 301030 301247 3014G4 30168130 302114 302331 302547 302764 302980:217 1 319G 3412 3628 3844 4059 4275 4491 4706 4921 5136 216 2 5351 5566 5781 5996 6211 6425 6639 6854 7068 7282 215 3 749G 7710 7924 8137 8351 8564 8778 8991 9204 9417 213 4 9G30 9843 310056 310268 310481 310693 310906 311118 311330 311542 212 5 311754 311966 2177 2389 2G00 2812 3023 3234 3445 3656 211 6 38G7 4078 4289 4499 4710 4920 5130 6340 5551 5760 210 7 5970 6180 6390 6599 6809 7018 7227 7436 7646 7854 209 8 80G3 8272 8481 8G89 8898 9106 9314 9522 9730 9938 208 9 320146 320354 3205G2|320709 320977 321184 321391)321598 321805 322012 207 210 322219 3224261322633 322839 32304G 323252 3234581323665 323871 324077 206 I 4282 4488 4G94 4899 5105 5310 6516 5721 5926 6131 205 2 6336 6541 G745 6950 7155 7359 7563 7767 7972 8176 204 3 8380 8583 8787 8991 9194 9398 9601 9805 330008 330211 203 4 330414 330617 330819 3?1022 331225 331427 331630331832 2034 2236 202 5 2438 2640, 2842 3044 3246 3447 3649, 3850 4051 4253 202 6 4454 4655| 4856 5057 5257 5458 5658 5859 6059 6260 201 7, 6460 6660, G8G0 70(0 7260 7459 7659 7858 8058 8257 200 8 845G 8650 3855 90541 9253 9451 9G50 1 98493400471340246 199 9 , 340444 340642 l 340841 3410391341237 341435 341G323418301 20281 22251138 N.| 1 2 3 | 4 |, i) 7 | 8 ■9 D. LOGARITHMS N. | 1 2 3 | 4 5 6 7 | 8 9 ID 220 342423 342020 312817 343014 343212 343409 343000 343&02 343999 344196 197 1 4392 4589 4785 4981 6178 5374 6570 5766 6962 6167 196 2 6353 6549 6744 6939 7135 7330 7525 7720 7915 8110 195 3 8305 8500 8694 8889 9083 9278 9472 9666 9860 350054 194 4 350248 350442 350636 350829 351G23 351216 351410 351(03 351796 1989 193 6 2183 2375 2568 2761 2954 3147 3339 3532 3724 3916 103 6 4108 4301 4493 4685 4876 6068 5260 6452 6643 5834 102 7 6026 6217 6408 6599 6790 6981 7172 7363 7554 7744 101 8 7935 8125 8316 8506 8696 8886 9076 9266 9456 9646 190 9 9835 360025 360215 360404 360593 360783 360972 361161 361350 361539 189 230 361728 361917 362105 362294 362482 362671 362859 363048 363236 363424 188 1 3612 3800 3988 4176 4363 4551 4739 4926 5113 5301 188 2 6488 6675 5862 6049 6236 6423 6610 6796 6983 7169 187 3 7356 7542 7729 7915 8101 8287 8473 8659 8845 9030 180 4 9216 9401 9587 9772 9958 370143 370328 370513 370098 370883 185 5 371068 371253 371437 371622 371806 1991 2175 2360 2544 2728 184 6 2912 3096 3280 3464 3647 3831 4015 4198 4382 4565 184 7 4748 4932 6115 6298 5481 5664 5846 6029 6212 6394 183 8 6577 6759 6942 7124 7306 7488 7670 7852 8034 8216 182 9 8398 8580 8761 8943 9124 9306 9487] 9068 9849 380030 181 240 380211 380392 380573 380754 380934 381115381296 381476 381056 381837 181 1 2017 2197 2377 2557' 2737 2917 3097 3277 3456 3636 180 2 3815 3995 4174 4353! 4533 4712 4891 6070 5249 5428 179 3 5606 6785 6964 6142 6321 6499 0077 6856 7034 7212 178 4 7390 7568 7746 7923 8101 8279 8456 8634 8811 8989 178 5 9166 9343 9520 9698 9875 300051 390228 390405 390582 390759 177 6 390935 391112 391288 391464 301641 1817 1993 2109 23451 2521 176 7 2697 2873 3048 3224 3400 3575 3751 3926 4101 4277 176 8 4452 4627 4802 4977 5152 5326 5501 6676 5850 6025 175 9 6199 6374 6548 6722 6896 7071 7245 7419 7592| 7766 174 250 397940 398114 398287 398461 398634 308808 398981,399154 399328 399501 173 1 9674 9847 400020 400192 400365 400538 4007111400883 401056 401228 173 2 401401 401573 1745 1917 2089 2261 24331 2605 2777 2949 172 3 3121 3292 3464 3635 3807 3978 4149 4320 4492 4663 171 4 4834 6005 6176 5346 6517 6688 5858 6029 6199 6370 171 5 6540 6710 6881 7051 72L1 7301 7561 7731 7901 8070 170 6 8240 8410 8579 8749 8918 9087 9257 9426 9595 9764 It. 9 7 9933 410102 410271 410440 410609 410777 410940 411114 411283 411451,169 ■ 8 411620 1788 1956 2124 2293 2461 2629 2796 2964 3132168 9 260 3300 414973 3467 3635 3803 31)70 4137 4305 4472| 4639 4806|l67 415140 415307 415474 415641 415808 415974 416141 410308,4: 1 6641 6807 6973 7139 7306 7472 7638 7804 7970! 8135|loo 2 8301 8467 8633 8798 8964 9129 9295 9400 962.'. 97911 10.) 3 9956 420121 420286 420451 420616 420781 420045 421110 421275 421439 l«o 4 421604 1768 1933 2097 2261 2426 2590 2754 2918 3082 104 5 3246 3410 3574 3737 3901 4065 4228 4392 4555 4718 164 6 4882 6045 5208 6371 6534 5697 5860 6023 6186 6349 163 7 6511 6674 6836 6999 7161 7324 7486 7648 7811 7973 162 8 8135 8297 8459 8621 8783 8944 9106 9268 0120 9591 162 9 9752 9914 430075 430236 430308 430559 430720 430881 431042 431203 Hil 270 431364 431525 431685 431846 432007 432167 432328 432488 432640 432809 161 1 2969 3130 3290 3450 3610 3770 3930 4090 4249i 4409 100 2 4569 4729 4888 6048 6207 6367 6526 6685 6844 6004 159 3 6163 6322 6481 6640 6799 6957 7116 7275 7433 75921159 4 7751 7909! 8067 8226 83X4 8542 870l] 8859 9017 9175ll58 6 9333 9491 9648 9806 9964 440122 440279440137 440594 440752; 158 6 4 40909 44106(14 11224 441381 441538 1605 1852 1 2009 2106 2323 157 7 2480 2637 2793 2950 3100 3203 3419 3576 3732 3889 167 8 4045 4201 4357 4513 4669 4825 4981 5137 6293 5449 156 9; 6604 6760 5915 6071 6226 6382 6537 6692 6848 7003 165 N | 0' 1 3 9 | D. OF NUMBERS. N.I 1 | 2 3 | 4 5 | 6 | 7 8 | 9 | D. 280 447158 447313 447408 447023 ( 147778 447933 448088 448242 4483 9' 7 148552 15.. 1 870C 8861 9015 9170 9324 9478 9633 9787 9941 450095 154 2 450249 450403 450557 450711 ' 150865 451018 451172 461326 451479 1633 154 3 1786 1940 2093 2247 2400 2553 2706 2859 3012 3165 153 4 3318 3471 3624 3777 3930 4082 4235 4387 4540 4692 153 5 4845 4997 5150 6302 5454 5606 5758 6910 6062 6214 152 6 6366 6518 6670 6821 6973 7125 7276 7428 7579 7731 152 7 7882 8033 8184 8336 8487 8638 8789 8940 9091 9242 151 8 9392 9543 9694 9845 9995 460146 460296 460447 460597 460748 151 9 460898 461048 461198 461348 161499 1649| 1799 1948 2098 2248 150 290 462398 462548 462697 462847 - 162997 463146 463296 463445 463594 463744 150 1 3893 4042 4191 4340 4490 4639 4788 4936 5085 6234 149 2 6383 6532 6680 5829 6977 6126 6274 6423 6571 6719 149 3 6868 7016 7164 7312 7460 7608 7756 7904 8052 8200 148 4 8347 8495 8643 8790 8938 9085 9233 9380 9527 9675 148 5 9822 9969 470116 470263 - 170410 470557 470704 470851 470998 471145 147 6 471292 471438 1685 1732 1878 2025 2171 2318 2464 2610 146 7 2756 2903 3049 3195 3341 3487 3633 3779 3925 4071 146 8 4216 4362 4508 4653 4799 4944 5090 5235 6381 6526 146 9 5671 5816 5962 6107 6252 6397 6542 6687 6832 6976 145 300 477121 477266 477411 477555 477700 477844 477989 478133 478278 478422.14 1 8566 8711 8855 8999 9143 9287 9431 9575 9719 9863 144 2 480007 480151 480294 480438 480582 480725 480869 481012 481156 481299 144 3 1143 1586 1729 1872 2016 2159 2302 2445 2588 2731 143 4 2874 3016 3159 3302 3445 3687 3730 3872 4015 4157 143 5 4300 4442 4585 4727 4869 5011 6153 5295 6437 6579 142 6 6721 6863 6005 6147 6289 6430 6572 6714 6855 6997 142 7 7138 7280 7421 7563 7704 7845 7986 8127 8269 8410 141 8 9 8551 9958 8692 490099 8833 490239 8974 490380 9114 490520 9255 9396 9537 490941 9677 491081 9818 491222 141 140 490661 490801 310 491362 491502 491642 491782 491922 492062 492201 492341 492481 492621 140 1 2760 2900 3040 3179 3319 3458 3597 3737 3876 4015 139 2 4155 4294 4433 4572 4711 4850 4989 6128 5267 5406 139 3 5544 5683 6822 6960 6099 6238 6376 6515 6653 6791 139 4 6930 7068 7206 7344 7483 7621 7759 7897 8035 8173 138 5 8311 8448 8586 8724 8862 8999 9137 9275 9412 9550 138 6 9687 9824 9962 500099 500236 500374 500511 500648 500785 500922 137 7 501059 501196 501333 1470 1607 1744 1880 2017 2154 2291 137 8 2427 2564 2700 2837 2973 3109 3246 3382 3518 3655 136 9 3791 3927 4063 4199 4335 4471 4607 4743 4878 5014 136 320 605150 505286 505421 505557 505693 505828 505964 50-6099 506234 506370 136 1 6505 6640 6776 6911 7046 7181 7316 7451 7586 7721 135 2 7856 7991 8126 8260 8395 8530 8664 8799 8934 9068 135 3 9203 9337 9471 9606 9740 9874 510009 510143 510277 510411 134 4 510545 510679 510813 510947 511081 511215 1349 1482 1616 1750 134 5 1883 2017 2151 2284 2418 2551 2684 2818 2951 3084 133 6 3218 3351 3484 3617 3750 3883 4016 4149 4282 4415 133 7 4548 4681 4813 4946 5079 5211 5344 5476 5609 5741 133 8 5874 6006 6139 6271 6403 6535 6668 6800 6932 7064 132 9 7196 7328 7460 7592 7724 7855 7987 8119 8251 8382 132 330 518514 518641 518777 518909 519040 519171 519303|ol9434 520615L520745 519566)519697 131 1 982S 9958 52009C 520221 520353 520484 520876)521007 131 5 52113$ 52126S 1400 1530 1661 1792 is*22 2053 2183 2314 131 a 2444 t 257J 2705 2835 2966 3096 3226 335fc 3486 .1616 130 4 . 374( > 387( > 400f 4136 4266 439(j 452C 465fc 4785 4915 130 £ 504J > 5174 l 5304 , 6434 6563 5693 6822 6951 6081 6210 129 ( 1 633! ) 646J > 659S ! 6727 6856 6985 7114 7243 7372 7501 129 I r 763( ) 7759,' 788* i 8016 8145 8274 8405 8531 8660 8788 12^ i i 891" f 90451 9174 L 9302 9430 9559 9681 981E 9943 530072 128 i ) 630200 530328|530456 530584 530712 530840 53096$ 53109t .531223 1351 m N. 1 2 3 4 || 5 6 7 8 9 6 LOGARITHMS TT 1 2 3 4 6 6 7 8 9 1). 340 531479 531GU7 531734 531862 531990 532117 532245 632372 532500 532627 128 l 2754 2882 3009 3136 3264 3391 3518 3645 3772 3899 127 2 402b 4153 4280 4407 4534 4661 4787 4914 5041 5167 127 3 5294 5421 5547 5674 6800 6927 6053 6180 6306 6432 126 4 6558 6685 6811 6937 7063 7189 7315 7441 7567 7693 126 5 7819 7945 8071 8197 8322 8448 8574 8699 8825 8951 126 6 9076 9202 9327 9452 9578 9703 9829 9954 540079 540204 125 7 540329 540455 540580 540705 540830 540955 541080 541205 1330 1454 125 8 1579 1704 1829 1953 2078 2203 2327 2452 2576 270] 125 9 2825 2950 3074 3199 3323 3447 3571 3696 3820 3944 124 350 544068 544192 544316 544440 544564 544688 544812 544936 545060 5451831: 1 5307 5431 5555 5678 6802 6925 6049 6172 6296 6419 124 2 6543 6666 6789 6913 7036 7159 7282 7405 7529 7652 123 3 7775 7898 8021 8144 8267 8389 8512 8635 8758 8881 123 4 9003 9126 9249 9371 9494 9616 9739 9861 9984|550106 123 5 550228 550351 550473 550595 550717 550840 550962 551084 551206 1328 122 6 1450 1572 1694 1816 1938 2060 2181 2303 2425 2547 122 7 2668 2790 2911 3033 3155 3276 3398 3519 3640 3762 121 8 3883 4004 4126 4247 4368 4489 4610 4731 4852 4973 121 9 5094 5215 5336 5457 6578 6699 5820 6940 6061 6182121 360 556303 556423 556544 656664 556785 556905 557026 557146 557267 557387 120 1 7507 7627 7748 7868 7988 8108 8228 8349 8469 8589 120 2 8709 8829 8948 9068 9188 9308 9428 9548 9667 9787 120 3 9907 060026 560146560265 560385 560504 560624 560743 560863 560982 119 4 561101 1221 1340 1459 1578 1698 1817 1936 2055 2174 119 5 2293 2412 2531 2650 2769 2887 3006 3125 3244 3362 119 6 3481 3600 3718 3837 3955 4074 4192 4311 4429 4548119 7 4666 4784 4903 5021 5139 5257 5376 5494 5612 5730118 8 5848 5966 6084 6202 6320 6437 6555 6673 6791 6909 118 9 7026 7144 7262 7379 7497 7614 7732 7849 7967 8084 lib 370 568202 568319 568436 568554 568671 568788 568905 569023 5691401569257 117 1 9374 9491 9608 9725 9842 9959 570076 570193 570309 570426 117 2 570543 570660 570776 570893 571010 571126 1243 1359 1476 1592 117 3 1709 1825 1942 2058 2174 2291 2407 2523 2639 2755 116 4 2872 2988 3104 3220 3336 3452 3568 3684 3800 3915 116 5 4031 4147 4263 4379 4494 4610 4726 4841 4957 5072 116 6 5188 6303 5419 5534 6650 6765 6880 6996 6111 6226 115 7 6341 6457 6572 6687 6802 6917 7032 7147 7262 7377 115 8 7492 7607 7722 7836 7951 8066 8181 8295 8410 8525 115 9 8639 8754 8868 8983 9097 9212 9326 9441 9555 9669 114 380 579784 579898 580012 5801261580241 580355 580469 580583 580697 580811 114 1 680925 581039 1153 1267 1381 1495 1608 1722 1836 1950 114 2 2063 2177 2201 2404 2518 2631 2745 2858 2972 3085 114 3 3199 3312 3 126 3539 3652 3765 3879 3992 4105 4218 m 4 4331 4444 4557 4670 4783 4896 6009 ' 5122 5235 5348 lit 5 5461 5574 5686 5799 5912 6024 6137 6250 6362 6475 113 6 6587 6700 6812 6925 7037 7149 7262 7374 7486 7599 112 7 7711 7823 7935 8047 8160 8272 8384 8496 8608 8720 112 8 8832 8944 9056 9167 9279 9391 9503 9615 9726 9838 112 9 9950r.«)0061 590173 590284 590396 590507 590619 590730 590842 590953 112 390 591065 591176 591287 591399 591510 591621 591732 591843 591955 592066J111 1 2177 2288 2399 2510 2621 2732 2843 2954 3064 3175illl 2 3286 3397 3508 3618 3729 3840 3950 4061 4171 4282llll 3 4393 4503 4614 4724 4834 4945 6055 5165 5276 6386110 4 6496 5606 5717 6827 5937 6047 6157 6267 6377 6487|110 5 6597 6707 6817 6927 7037 7146 7256 7366 7476 7586 110 6 7695 7805 7914 8024 8134 8243 8353 8462 8572 8681 110 7 8791 8900 9009 9119 9228 9337 9446 9556 9665 9774 109 8 9833 9992 600101 600210 600319 600428600537 600646 600755 600864 109 9 600973 601082 1191 1299 1408 1517 1625 1734 1843| 1951109 N.| 1 2 * 3 5 6 7 8 B ID. OF NUMBERS. 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UF NUMBERS. 9 N. 1 I D. 520 I 2 3 4 5 6 7 8 9 716003 6838 7671 8502 9331 720159 0986 1811 2634 3456 716087 6924 7754 8585 9414 720242 1068 1893 2716 3538 7L6170 7004 7837 8668 9497 720325 1151 1975 2798 3620 716254 716337 7088 71711 7920 8003 8751 8834 9580 9663 720407 720490 1233 1316 2058 2140 2881 2963 3702 3784 716421 7254 8086 8917 9745 720573 1398 2222 3045 3866 7338 8169 9000 9828 720655 1481 2305 3127 3948 530,724276 5095 5912 6727 7541 8354 9165 9974 8 730782 9 1589 540 1 2 3 4 5 6 7 8 9 724358 5176 5993 1 6809 7623 8435 9246 730055 0863 1669 724440 5258 6075 6890 7704! 85161 9327 730136 0944 1750 724522 5340 6156 6972 7785 8597 9408 730217 1024 1830 724601 5422 6238 7053 7866 8678 9489 730298 1105 1911 732394 3197 3999 4800 6599 6397 7193 7987 8781 9572 732474 3278 4079 4880 6679 6476 7272 8067 8860 9651 732555 3358 4160 4960 6759 6556 7352 8146 8939 9731 732635 3438 4240 5040 5838 6635 7431 8225 9018 9810 732715 3518 4320 5120 6918 6715 7511 8305 9097 9889 550 740363 1 1152 1 1939 3 2725 4 3510 5 4293 6 5075; 7 58551 8 6034 ! 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OF NUMBERS. 11 N. 1 2 3 4 6 6 7 8 9 | D. 640 806180 806248 806316 806384 806451 8065 li> 806587 806655 806723 806790 68 1 6858 6926 6994 7061 7129 7197 7264 7332 7400 7467 68 2 7535 7603 7670 7738 7806 7873 7941 8008 8076 8143 68 3 8211 8279 8346 8414 8481 8549 8616 8684 8751 8818 67 4 8886 8953 9021 9088 9156 9223 9290 9358 0425 9492 67 5 9560 9627 9694 9762 9829 9896 9964 810031 810098 810165 67 6 810233 810300 810367 810434 810501 810569 810636 0703 0770 0837 67 7 0904 0971 1039 1106 1173 1240 1307 1374 1441 1608 67 8 1575 1642 1709 1776 1843 1910 1977 2044 2111 2178 67 9 2245 2312 2379 2445 2512 2579 2646 2713 2780 2847| 67 (150 812913 812980 813047 813114 813181 813247 813314 813381 813448 8] 67 1 3581 3648 3714 3781 3848 3914 3981 4048 4114 4181 67 2 4248 4314 4381 4447 4514 4581 4647 4714 4780 4847 67 3 4913 4980 5046 5113 5179 5246 5312 6378 6445 6511 66 4 6678 5644 6711 5777 5843 5910 6976 6042 6109 6175 66 6 6241 6308 6374 6440 6506 6573 6639 6705 6771 6838 66 6 6904 6970 7036 7102 7169 7235 7301 7367 7433 7499 66 7 7565 7631 7698 7764 7830 7896 7962 8028 8094 8160 66 8 8226 8292 8358 8424 8490 8556 8622 8688 8754 8820 66 9 8885 8951 9017 9083 9149 9215 9281 9346 9412 9478 66 660 819544181 819676 819741 819807 819873 819939 820004 820070 820136 66 1 820201 820267 820333 820399 820464 820530 820595 0661 0727 0792 66 2 0858 0924 0989 1055 1120 1186 1251 1317 1382 1448 66 3 1514 1579 1645 1710 1775 1841 1906 1972 2037 2103 65 4 2168 2233 2299 2364 2430 2495 2560 2626 2691 2756 65 5 2822 2887 2952 3018 3083 3148 3213 3279 3344 3409 66 6 3474 3539 3605 3670 3735 3800 3865 3930 3996 4061 65 7 4126 4191 4256 4321 4386 4451 4616 4581 4646 4711 65 8 4776 4841 4906 4971 5036 5101 5166 5231 5296 5361 65 9 6426 5491 5556 5621 5686 5751 5815 6880 6945 6010 65 670 826075 826140 8: 826269 826334 826399 826464 826528 826593 826658 65 1 6723 6787 6852 6917 6981 7046 7111 7175 7240 7305 66 2 7369 7434 7499 7563 7628 7692 7767 7821 7886 7951 65 3 8015 8080 8144 8209 8273 8338 8402 8467 8531 8595 64 4 8660 8724 8789 8853 8913 8982 9046 9111 9175 9239 64 5 9304 9368 9432 9497 9561 9625 9690 9754 9818 9882 64 6 9947 830011 830075 830139 830204 830268 830332 830396 830460 830525 64 7 830589 0653 0717 0781 0845 0909 0973 1037 1102 1166 64 8 1230 1294 1358 1422 1486 1550 1614 1678 1742 1806 64 9 1870 1934 1998 2062 2126 2189 2253 2317 2381 2446 64 680 832509 832573 832637 832700 83 832828 832892 832956 833020 8! 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X 2 3 4 1 5 6 7 8 9 | D. 765 880814 880871 880928 880985 881042 881099 881156 881213 881271 8813281 67 1 1385 1442 1499 1556 1613 1670 1727 1784 1841 1898 67 2 1955 2012 2069 2126 2183 2240 2297 2354 2411 2468 67 3 2525 2581 2638 2695 2752 2809 2866 2923 2980 3037 67 4 3093 3150 3207 3264 3321 3377 3434 3491 3548 3605 57 5 3661 3718 3775 3832 3888 3945 4002 4059 4115 4172 67 6 4229 4285 4342 4399 4455 4512 4569 4625 4682 4739 57 7 4795 485'2 4909 4965 6022 5078 6135 5192 5248 6305 57 8 5361 6418 5474 6531 5587 6644 6700 6757 5813 6870 57 9 6926 0383 6039 6096 6152 6209 6265 6321 6378 6434 66 770 886491 88054»7 88<A504 886660 886716 886773 880829.88 886942 «86fi<9S 66 1 7054 7111 7167 7223 7280. 7336 7392 744i> 7505 7561 56 2 7617 7674 7730 7786 7842. 7898 7955 8011 8067 8123 56 3 8179 8236 8292 8348 8404 8460 8516 8573 8629 8685 56 4 8741 8797 8853 8909 8965 9021 9077 9134 9190 9246 56 5 9302 9358 9414 9470 9526 9582 9638 9691 9750 9806 56 6 9862 9918 9974 890030 890086 890141- 890197 890253 890309 890365 56 7 890421 890477 890533 0589 0645 0700 0756 0812 0808 0924 56 8 0980 1035 1091 1147 1203 1259 1314 1370 1426 1482 66 9 1537 1593 1649 1705 1760 1816 1872 1928 1983 2039 56 780 892095 892150 892206 8i 892317 892373 8! 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OF NUMBERS 15 gj o 1 2 3 4 3 6 7 8 9 TT 880 944483 944532 944581 944631 944680 944729 944779 944828 944877 944927 49 1 4976 5025 5074 6124 5173 5222 6272 5321 6370 64J 9 49 2 5469 5518 5567 5616 5665 5715 5764 5813 6862 69 v 12 49 3 5961 6010 6059 6108 6157 6207 6256 6305 6354 6403 49 4 6452 6501 6551 6600 6649 6698 6747 6796 6845 6894 49 6 6943 6992 7041 7090 7140 7189 7238 7287 7336 7385 49 6 7434 7483 7532 7561 7630 7679 7728 7777 7826 7875 49 7 7924 7973 8022 8070 8119 8168 8217 8266 8315 8364 49 8 8413 8462 8511 8560 8609 8657 8706 8755 8804 8853 49 9 8902 8951 8999 9048 9097 9146 9195] 9244 9292 9341 49 890 949390 949439 949488 949536 949585 949634 949683 949731 949780 949829 49 1 9878 9926 9975 950024 950073 950121 950170 950219 950267 950316 49 2 950365 950414 950462 0511 0560 0608 0657 0706 0754 0803 49 3 0851 0900 0949 0997 1046 1095 1143 1192 1240 1289 49 4 1338 1386 1435 1483 1532 1580 1629 1677 1726 1775 49 5 1823 1872 1920 1969 2017 2066 2114 2163 2211 2260 48 6 2308 2356 2405 2453 2502 2550 2599 2647 2696 2744 48 7 2792 2841 2889 2938 2986 3034 3083 3131 3180 3228 48 8 3276 3325 3373 3421 3470 3518 3566 3615 3663 3711 48 9 3760 3808 3856 3905 3953 4001 4049 4098 4146 4194 48 900 954243 9542? 1 954339 95438 954435 954484 954532 954580 954628 954677 48 1 4725 4773 4821 4869 4918 4966 6014 5062 5110 5158 48 2 5207 5255 6303 5351 5399 5447 6495 5543 6592 5640 48 3 5688 5736 5784 5832 6880 6928 5976 6024 6072 6120 48 4 6168 6216 6265 6313 6361 6409 6457 6505 6553 6601 48 5 6649 6697 6745 6793 6840 6888 6936 6984 7032 7080 48 6 7128 7176 7224 7272 7320 7368 7416 7464 7512 7559 48 7 7607 7655 7703 7751 7799 7847 7894 7942 7990 8038 48 8 8086 8134 8181 8229 8277 8325 8373 8421 8468 8516 48 9 8564 8612 8659 8707 8755 8803 8850 8898 8946 8994 48 910 959041 959089 959137 959185 959232 959280 959328 959375 959423 959471 48 1 9518 9566 9614 9661 9709 9757 9804 9852 9900 9947 48 2 9995 960042 960090 960138 960185 960233 960281 960328 960376 960423 48 3 960471 0518 0566 0613 066L 0709 0756 0804 0851 0899 48 4 0946 0994 1041 1089 1136 1184 1231 1279 1326 1374 48 5 1421 1469 1516 1563 1611 1658 1706 1753 1801 1848 47 6 1895 1943 1990 2038 2085 2132 2180 2227 2275 2322 47 7 2369 2417 2464 2511 2559 2606 2653 2701 2748 2795 47 8 2843 2890 2937 2985 3032 3079 3126 3174 3221 3268 47 9 3316 3363 3410 3457 3504 3552 3599 3646 3693 3741 47 920 963788 963835 963882 963929 963977 964024 964071 964118 964165 964212 47 1 4260 4307 4354 4401 4448 4495 4542 4590 4637 4684 47 2 4731 4778 4825 4872 4919 4966 6013 5061 6108 5155 47 3 5202 6249 5296 5343 5390 5437 5484 5531 6578 6625 47 4 5672 6719 5766 6813 6860 5907 5954 6001 6048 6095 47 5 6142 6189 6236 6283 6329 6376 6423 6470 6517 6564 47 6 6611 6658 6705 6752 6799 6845 6892 6939 C986 7033 47 7 7080 7127 7173 7220 7267 7314 7361 7408 7454 7501 47 8 7548 7595 7642 7688 7735 7782 7829 7875 7922 7969 47 9 8016 8062 8109 8156 8203 6249 8296 8343 8390 8436 47 930 968483 968531 968576 968623 968670 968716 968763 96881 968856 968903 47 1 8950 8996 9043 9090 9136 9183 9229 9276 9323 9369 47 2 9416 9463 9509 9556 9602 9649 9695 9742 9789 9835 47 3 9882 9928 9975 970021 970068 970114 970161 970207 970254 970300 47 4 970347 970393 970440 0486 0533 0579 0626 0672 0719 0765 46 5 0812 0858 0904 0951 0997 1044 1090 1137 1183 1229 46 6 1276 1322 1369 1415 1461 1508 1554 1601 1647 1693 46 7 1740 1786 1832 1879 1925 1971 2018 2064 2110 2157 46 8 2203 2249 2295 2342 2388 2434 2481 2527 2573 2619 46 9 2666 2712 2758 2804 2851 2897 2943 2989 3035 3082 46 N. 1 4 5 « 9 .d. 16 LOGARITHMS OF NUMBERS. 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